Work, Energy, and Power - Class 11 Physics - Chapter 6 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Work, Energy, and Power | NCERT | Physics | Class 11
A person carrying a bag of total mass $25 \mathrm{~kg}$ climbs up to a height of $5 \mathrm{~m}$ in 30 seconds slowly. The power delivered for carrying the bag is:
A) $40.83 \mathrm{~W}$
B) $19.67 \mathrm{~W}$
C) $26.23 \mathrm{~W}$
D) $38.15 \mathrm{~W}$
The correct answer is Option A: $40.83 \mathrm{~W}$.
To solve this problem, we'll start by calculating the force exerted by the person, which is equal to the weight of the bag. This force ($F$) is given by: $$ F = mg = 25 , \text{kg} \times 9.8 , \text{m/s}^2 = 245 , \text{N} $$
Next, we calculate the work done ($W$) by this force to lift the bag to a height of $5 , \text{m}$: $$ W = F \times h = 245 , \text{N} \times 5 , \text{m} = 1225 , \text{J} $$
Since the work is done over a period of 30 seconds, we can find the power delivered ($P$) using the formula: $$ P = \frac{W}{t} = \frac{1225 , \text{J}}{30 , \text{s}} = 40.83 , \text{W} $$
Thus, the power delivered for carrying the bag is $40.83 \mathrm{~W}$.
The commercial unit of energy consumption in households, industries, and commercial establishments is:
A. $\mathrm{kW}$
B. $\mathrm{kWh}$
C. Joule
D. Watt
The correct answer is B. $\mathrm{kWh}$
$\mathrm{kWh}$, kilowatt-hour, is the standard unit of energy consumption used across households, industries, and commercial establishments. This unit measures the energy used over a period of time, providing a practical quantification for billing and energy management purposes.
Which of the following statements is true regarding hydropower plants?
A. The potential energy possessed by the stored water is converted into electricity.
B. Kinetic energy possessed by the stored water is converted into potential energy.
C. Electricity is extracted from the water.
D. Water is converted into steam to produce electricity.
The correct answer is A. The potential energy possessed by the stored water is converted into electricity.
In a hydropower plant, water stored at a higher elevation has significant potential energy due to its position. This water is released to flow down through a penstock pipe, transitioning its energy form from potential to kinetic as it moves towards the turbines. This kinetic energy is then utilized to rotate the turbine blades, thereby converting it into mechanical energy. This mechanical energy drives the generators to convert it finally into electrical energy, which is then distributed for use. Thus, the sequence of energy transformation in a hydropower plant is as follows: $$ \text{Potential Energy} \rightarrow \text{Kinetic Energy} \rightarrow \text{Mechanical Energy} \rightarrow \text{Electrical Energy}. $$
What is the type of energy that we get directly from the sun?
A. Solar energy B. Kinetic energy C. Potential energy
The correct answer is A. Solar energy.
Solar energy refers to the radiant light and heat from the Sun that can be harnessed using various evolving technologies, such as solar heating, solar thermal energy, photovoltaics, molten salt power plants, and artificial photosynthesis. It serves as a crucial source of renewable energy.
Define the term power. State its SI unit.
Power is defined as the rate at which work is done or energy is transferred. The SI unit of power is the watt (W).
A 1000 W oven is used 60 minutes every day. Calculate the number of units of electrical energy the oven consumes in 30 days.
To solve this problem, first, understand that power is calculated as the rate at which energy is used (or work is done). The formula for power can be expressed as: $$ P = \frac{E}{t} $$ where $P$ is power in watts, $E$ is energy in joules, and $t$ is time in seconds.
Given:
The power of the oven, $P = 1000 , \text{W}$
The oven is used for $60 , \text{minutes}$ per day. In seconds, this is: $$ 60 , \text{minutes} \times 60 , \text{seconds/minute} = 3600 , \text{seconds} $$
Now substitute these values into the power formula to find the energy used per day: $$ 1000 , \text{W} = \frac{E}{3600 , \text{s}} $$ Solving for $E$ gives: $$ E = 1000 , \text{W} \times 3600 , \text{s} = 3.6 \times 10^6 , \text{joules} $$ This is the energy consumed in one day.
We know that one unit of electrical energy is equivalent to $3.6 \times 10^6$ joules. Therefore, the number of units of energy the oven uses in one day is: $$ \frac{3.6 \times 10^6 , \text{joules}}{3.6 \times 10^6 , \text{joules/unit}} = 1 , \text{unit} $$
Finally, to find the total energy consumed over 30 days: $$ 1 , \text{unit/day} \times 30 , \text{days} = 30 , \text{units} $$
Thus, the oven consumes 30 units of electrical energy in 30 days.
A non-conservative force dissipates energy, while a conservative force does not dissipate energy.
A) True
B) False
The correct answer is A) True.
Conservative forces, such as gravity and spring forces, do not dissipate energy; they are capable of converting energy entirely between potential and kinetic forms without any loss.
However, non-conservative forces lead to energy dissipation, typically converting mechanical energy into thermal energy, such as friction or air resistance. This energy cannot be fully recovered in a simple mechanical process.
Write the energy conversion that takes place in a hydropower plant.
In a hydropower plant, the stored water in the dam possesses potential energy due to its elevation. As the water flows downward from this higher altitude, this potential energy transforms into kinetic energy. This flowing water drives the turbine, converting the kinetic energy into mechanical energy. Subsequently, the generator in the plant converts this mechanical energy into electrical energy, which can then be utilized for various purposes.
An unruly demonstrator lifts a stone of mass $200 \mathrm{~g}$ from the ground and throws it at his opponent. At the time of projection, the stone is $150 \mathrm{~cm}$ above the ground and has a speed of $3.00 \mathrm{~m} / \mathrm{s}$. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw, what horsepower does he use?
The given data is:
Mass of the stone, $ m = 200 \text{ g} = 0.2 \text{ kg} $,
Height from the ground, $ h = 150 \text{ cm} = 1.5 \text{ m} $,
Speed of the stone, $ v = 3.00 \text{ m/s} $,
Time for the action, $ t = 1 \text{ sec} $.
The total work done during the process of lifting and throwing the stone can be calculated using the formula for mechanical work which accounts for both kinetic energy and potential energy: $$ \text{Total work done} = \frac{1}{2} mv^2 + mgh $$
Substituting the values, we get: $$ \frac{1}{2} \cdot 0.2 \cdot (3.00)^2 + 0.2 \cdot 9.8 \cdot 1.5 = 0.09 + 2.94 = 3.03 \text{ J} $$
To find the horsepower used, we convert the work done from joules to horsepower-hours. The conversion factor is that $1 \text{ hp} = 746 \text{ W}$, and power is work done per unit time: $$ \text{Power (in watts)} = \frac{\text{Work done (in joules)}}{\text{Time (in seconds)}} = \frac{3.03}{1} = 3.03 \text{ W} $$
Converting watts to horsepower: $$ \text{Power (in horsepower)} = \frac{3.03 \text{ W}}{746} = 0.00406 \text{ hp} $$
In summary:
Total work done: $3.03 \text{ J}$
Horsepower used: $0.00406 \text{ hp}$
Is it possible to find the electrical energy in terms of $\mathrm{kJ}$ if the power consumed by a toaster in 15 min is 125 $\mathrm{W}$?
A) 112.5 $\mathrm{kJ}$
B) 112.05 $\mathrm{kJ}$
C) 11.25 $\mathrm{kJ}$
D) 1125 $\mathrm{kJ}$
The correct option is A$$ 112.5 \text{ kJ} $$
To calculate the electrical energy used by the toaster in terms of $\text{kJ}$, we start by converting the time:
Given: Time $t = 15$ minutes, so: $$ t = 15 \times 60 = 900 \text{ s} $$
The power $P$ consumed by the toaster is given as: $$ P = 125 \text{ W} $$
Using the formula for energy ($\text{H}$), where $\text{H} = P \cdot t$, we can calculate the total energy consumed during this period: $$ \text{H} = 125 \text{ W} \times 900 \text{ s} = 112500 \text{ J} $$
Since $1 \text{ kJ} = 1000 \text{ J}$, to convert joules to kilojoules: $$ \frac{112500}{1000} = 112.5 \text{ kJ} $$
Therefore, the energy consumed by the toaster in $\text{kJ}$ is 112.5 kJ. (Answer)
Which of the following situations involves a person using a simple machine to get work done?
A) A person sleeping on the couch.
B) A girl cutting an apple with a knife.
C) A man using sugar tongs to pick up sugar cubes.
D) A woman using a pulley to lift up a bucket of water from the well.
The correct answers where a simple machine is utilized to perform work are:
B) A girl cutting an apple with a knife
A knife acts as a wedge, which is a type of simple machine that helps split objects apart. Hence, it's used here to cut through the apple.
C) A man using sugar tongs to pick up sugar cubes
Sugar tongs function similarly to a lever, another form of simple machine. They allow easier lifting and maneuvering of small objects like sugar cubes through mechanical advantage.
D) A woman using a pulley to lift up a bucket of water from the well
The pulley system, a classic example of a simple machine, significantly reduces the effort needed to lift heavy loads vertically. It does so by distributing the weight and providing mechanical advantage.
The scenario A) A person sleeping on the couch does not involve any simple machines, as no mechanical work is being performed.
Which of the following define the work most accurately?
Work is the product of displacement and component of force in the direction of displacement.
Work is the product of force and displacement.
Work is the product of force and component of displacement in the direction of force.
Work is the product of distance and force applied.
A) 1 and 3
B) 2 and 4
C) 1, 2, and 3
D) All of these
The correct answer is A) 1 and 3.
Work, in physics, is defined in specific terms related to force and displacement. Let's examine the accuracy of the statements given:
Work is the product of displacement and component of force in the direction of displacement. This is correct. When a force acts on an object, only the component of the force that is parallel to the displacement contributes to the work done.
Work is the product of force and displacement. This statement simplifies the conditions under which work is calculated. It fails to consider that only the component of the force in the direction of the displacement is relevant.
Work is the product of force and component of displacement in the direction of force. This too is correct. In scenarios where the displacement isn't aligned with the force, only the component of displacement in the direction of the force is considered when calculating work.
Work is the product of distance and force applied. This statement is incorrect because it takes into account the total distance traveled rather than just the displacement (which is a vectored quantity), and it does not consider the direction of the force relative to the displacement.
Thus, statements 1 and 3 most accurately define the concept of work in physics.
Complete the following sentence: The conversion of part of energy into an undesirable form is called:
The conversion of part of energy into an undesirable form is called degradation of energy.
Which type of lever has the length of the effort arm greater than the length of the load arm?
A. Energy multiplier
B. Force multiplier
C. Speed multiplier
D. Distance multiplier
The correct answer is B. Force multiplier.
In a force multiplier lever, the length of the effort arm is greater than the length of the load arm. This setup allows the lever to multiply the force exerted, making it easier to lift heavier loads. Such levers are pivotal in mechanics where significant weight needs to be moved with less force.
The following are not included in internal energy:
Nuclear energy
Rotational energy
Vibrational energy
Energy arising from gravitational attraction
The correct answer is D.
The formula for internal energy is given by: $$ E = E_{\text{ele}} + E_{\text{nucl}} + E_{\text{chemical}} + E_{\text{pot}} + E_{\text{kinetic}} \left( E_t + E_v + E_r \right) $$ where:
$ E_{\text{ele}} $: Electrical energy
$ E_{\text{nucl}} $: Nuclear energy
$ E_{\text{chemical}} $: Chemical energy
$ E_{\text{pot}} $: Potential energy
$ E_{\text{kinetic}} $: Kinetic energy, which includes:
$ E_t $: Translational energy
$ E_v $: Vibrational energy
$ E_r $: Rotational energy
Gravitational energy is not included in the internal energy, making option D the correct answer.
A body of mass $m$ is moving with velocity $v$. If $P$ is the momentum associated with the body and K.E. is the kinetic energy, then K.E. and $P$ are related as:
A) K.E. $= \frac{P^{2}}{2m}$ B) K.E. $= \frac{P^{2}}{m}$ C) K.E. $= \frac{P^{2}}{2}$ D) K.E. $= \frac{P}{m^{2}}$
The correct option is $\mathbf{A}$.
To show how kinetic energy (K.E.) and momentum ($P$) are related, let's start with the formulas:
Kinetic Energy (K.E.):$$ \text{K.E.} = \frac{1}{2} mv^2 $$
Momentum (P):$$ P = mv $$
To establish the relationship between K.E. and $P$, we can manipulate the formula for kinetic energy by incorporating the expression for momentum.
Starting from the kinetic energy equation: $$ \text{K.E.} = \frac{1}{2} mv^2 \quad \text{(i)} $$
Using the momentum formula: $$ P = mv \quad \text{(ii)} $$
Next, we need to substitute $v$ from equation (ii) into the kinetic energy formula.
From equation (ii): $$ v = \frac{P}{m} $$
Substituting $ v^2 = \left(\frac{P}{m}\right)^2 $ into equation (i), we get: $$ \text{K.E.} = \frac{1}{2} m \left(\frac{P}{m}\right)^2 $$
Simplifying this expression: $$ \text{K.E.} = \frac{1}{2} m \cdot \frac{P^2}{m^2} $$
Resulting in: $$ \text{K.E.} = \frac{P^2}{2m} $$
Therefore, we can confirm that: $$ \text{K.E.} = \frac{P^2}{2m} $$
Thus, the correct option is $\mathbf{A}$.
In the shown pulley-block system, strings are light. Pulleys are massless and smooth and system is released from rest. In 0.6 second, what is the work done by gravity on 20 kg and 10 kg block respectively? Take $[g=10 \mathrm{~m/s^2}]$
A 120 J; 30 J B 30 J; 120 J C 240 J; -60 J D -240 J; 60 J
First, we analyze the forces acting on the blocks and determine the acceleration.
Free-Body Diagram (FBD) analysis:
For the $20 \text{ kg}$ block: [ 30g = 90a ] [ a = \frac{g}{3} ]
Displacement Calculation:
Next, we calculate the displacement of the $20 \text{ kg}$ block. Using the equation for displacement in uniformly accelerated motion: [ \text{Displacement of } 20 \text{ kg} \text{ block} = \frac{1}{2} a t^2 ] Since $a = \frac{2g}{3}$ and $t = 0.6 \text{ seconds}$: [ = \frac{1}{2} \times \frac{2g}{3} \times 0.6^2 ] [ = \frac{1}{2} \times \frac{2 \times 10 \text{ m/s}^2}{3} \times 0.36 ] [ = 1.2 \text{ m} ]
Due to the pulley system configuration, the displacement of the $10 \text{ kg}$ block is half of that of the $20 \text{ kg}$ block: [ \text{Displacement of } 10 \text{ kg} \text{ block} = \frac{1.2 m}{2} = 0.6 \text{ m} ]
Work Done by Gravity on Each Block:
For the $20 \text{ kg}$ block:[ W_{20} = mgh ] [ = 20 \times 10 \times 1.2 ] [ = 240 \text{ J} ]
For the $10 \text{ kg}$ block (moving upward):[ W_{10} = -mgh ] [ = -10 \times 10 \times 0.6 ] [ = -60 \text{ J} ]
Thus, the work done by gravity on the $20 \text{ kg}$ block is 240 J, and on the $10 \text{ kg}$ block, it is -60 J.
Hence, option C (240 J; -60 J) is the correct answer.
Can a body process energy even when it is not in motion? Explain your answer with an example.
Yes, a body can indeed possess energy even when it is not in motion. This type of energy is known as Potential Energy, which is due to the body’s position or configuration.
A common example of potential energy is gravitational potential energy. The formula for gravitational potential energy is given by:
$$ P = mgh $$
where:
m represents the mass of the body,
g is the acceleration due to gravity,
h is the height at which the body is positioned above the reference point.
For instance, consider a rock placed on top of a hill. Even though the rock is not moving, it has energy due to its elevated position. If the rock were to fall, this stored potential energy would convert into kinetic energy as the rock gains speed. This illustrates that potential energy is the energy stored in the body due to its position, which can be converted into other forms of energy when the body is set in motion.
In a bulb, electrical energy is converted into:
A. potential energy
B. kinetic energy
C. light energy
Correct Option: C - Light Energy
When an electric current passes through the filament of a bulb, typically made of tungsten, it causes the filament to heat up due to its high resistance. As a result, the filament emits visible light along with significant amounts of heat. Therefore, in a bulb, electrical energy is transformed primarily into both heat energy and light energy.
Work done by car is 250000 J in 10 seconds. Then power developed by the engine is ___ kW.
Given:
Work done ( = 250000 , \text{J} )
Time taken ( = 10 , \text{s} )
To find:
Power developed by the engine in kW.
:
Power ( (P) ) is calculated using the formula:
[ P = \frac{\text{work done}}{\text{time}} ]
Substituting the given values:
[ P = \frac{250000 , \text{J}}{10 , \text{s}} ]
Simplifying this expression:
[ P = 25000 , \text{W} ]
Since ( 1 , \text{kW} = 1000 , \text{W} ), we convert the power from watts to kilowatts:
[ P = \frac{25000 , \text{W}}{1000} = 25 , \text{kW} ]
Thus, the power developed by the engine is 25 kW.
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Ask Chatterbot AINCERT Solutions - Work, Energy, and Power | NCERT | Physics | Class 11
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
When a man lifts a bucket, he applies an upward force to counteract the gravitational force acting downward. Since the direction of the displacement (upwards) is the same as the direction of the applied force, the work done by the man is positive.
(b) Work done by gravitational force in the above case
In the previous case, the gravitational force acts downward while the bucket is moving upward. Since the force (downward) is in the opposite direction to the displacement (upward), the work done by the gravitational force is negative.
(c) Work done by friction on a body sliding down an inclined plane
Friction always opposes the direction of motion. If a body slides down an inclined plane, friction acts upward along the plane while the displacement of the body is downward. Therefore, the work done by friction is negative.
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity
For a body moving with a uniform velocity on a rough horizontal plane, the applied force is used to counteract the frictional force and maintain uniform motion. Since the direction of the applied force is the same as the direction of displacement, the work done by the applied force is positive.
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest
The resistive force of air (air drag) acts in the opposite direction to the motion of the pendulum. As the pendulum vibrates, air resistance acts in a direction opposite to its displacement, removing kinetic energy from the system and eventually bringing it to rest. Thus, the work done by the resistive force of air is negative.
Here is a summarized table:
Quantity | Work Done |
---|---|
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket | Positive |
(b) Work done by gravitational force in the above case | Negative |
(c) Work done by friction on a body sliding down an inclined plane | Negative |
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity | Positive |
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest | Negative |
A body of mass $2 \mathrm{~kg}$ initially at rest moves under the action of an applied horizontal force of $7 \mathrm{~N}$ on a table with coefficient of kinetic friction $=0.1$. Compute the
(a) work done by the applied force in $10 \mathrm{~s}$,
(b) work done by friction in $10 \mathrm{~s}$,
(c) work done by the net force on the body in $10 \mathrm{~s}$,
(d) change in kinetic energy of the body in $10 \mathrm{~s}$, and interpret your results.
Step-by-Step Solution:
Firstly, let us calculate the components needed one by one.
(a) Work done by the applied force in $10 , \mathrm{s}$
Net acceleration of the body:
Applied force $ F $ = $ 7 , \mathrm{N} $
Mass $ m $ = $ 2 , \mathrm{kg} $
Coefficient of kinetic friction $ \mu_k $ = $0.1 $
Gravitational acceleration $g \approx 9.81 , \mathrm{m/s^2}$
Friction force $ f $ = $ \mu_k \times m \times g $
Force of friction: $$ f = 0.1 \times 2 , \mathrm{kg} \times 9.81 , \mathrm{m/s^2} = 1.962 , \mathrm{N} $$
Net force: $$ F_{net} = 7 , \mathrm{N} - 1.962 , \mathrm{N} = 5.038 , \mathrm{N} $$
Thus, acceleration ( a ) is: $$ a = \frac{F_{net}}{m} = \frac{5.038 , \mathrm{N}}{2 , \mathrm{kg}} = 2.519 , \mathrm{m/s^2} $$
Distance covered in $ 10 , \mathrm{s} $:$$ s = \frac{1}{2} a t^2 = \frac{1}{2} \times 2.519 , \mathrm{m/s^2} \times (10 , \mathrm{s})^2 = 125.95 , \mathrm{m} $$
Work done by the applied force:$$ W = F \times s = 7 , \mathrm{N} \times 125.95 , \mathrm{m} = 881.65 , \mathrm{J} $$
(b) Work done by friction in $ 10 , \mathrm{s}$
Friction force $ f $ = $1.962 , \mathrm{N} $
Distance covered $ s $ = $ 125.95 , \mathrm{m} $
$$ W_{f} = - f \times s = -1.962 , \mathrm{N} \times 125.95 , \mathrm{m} = -247.08 , \mathrm{J} $$
(c) Work done by the net force on the body in $ 10 , \mathrm{s}$
Net force $ F_{net}$ = $5.038 , \mathrm{N} $
Distance covered $ s $ = ($ 125.95 , \mathrm{m} $
$$ W_{net} = F_{net} \times s = 5.038 , \mathrm{N} \times 125.95 , \mathrm{m} = 634.57 , \mathrm{J} $$
(d) Change in kinetic energy of the body in $ 10 , \mathrm{s}$
Final velocity after ( 10 , \mathrm{s} ):$$ v = a \times t = 2.519 , \mathrm{m/s^2} \times 10 , \mathrm{s} = 25.19 , \mathrm{m/s} $$
Change in kinetic energy:$$ \Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 2 , \mathrm{kg} \times (25.19 , \mathrm{m/s})^2 = 634.57 , \mathrm{J} $$
Interpretation:
Work Done by Applied Force: The applied force does positive work, as it is in the direction of the displacement.
Work Done by Friction: The friction force does negative work, as it opposes the motion.
Work Done by Net Force: This is the sum of the work done by the applied force and the work done by the friction force and is equal to the change in kinetic energy of the body.
Change in Kinetic Energy: The kinetic energy of the body increases by $634.57 , \mathrm{J}$ in ( 10 , \mathrm{s} ), which matches the work done by the net force. This is in agreement with the work-energy theorem.
Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Fig. 5.11
Let's analyze the given potential energy functions in Fig. 5.11 and identify the regions in which the particle cannot be found for the given energy (E). We will also determine the minimum total energy that the particle must have in each case and think of simple physical contexts for these potential energy shapes.
(a) Step Potential
Potential Energy Function:
$ V(x) = 0 $ for $ x < a $
$ V(x) = V_0 $ for $ x \geq a $
Analysis:
The particle cannot be found in the region $ x \geq a $ if $ E < V_0 $, as the potential energy exceeds the total energy.
The minimum total energy (E) the particle must have to be found in the region $ x \geq a$ is $ V_0 $.
Physical Context:
This resembles a particle encountering a potential barrier or a particle in a potential step, such as an electron hitting a potential barrier in a semiconductor.
(b) Piecewise Linear Potential
Potential Energy Function:
$ V(x) = V_0 $for $ x < a $
$ V(x) $ increases linearly for $ a \leq x < b $ reaching $ V_3 $
$ V(x) = V_3 $ for $ b \leq x < d $
$ V(x) = V_0 $ for $ x \geq d $
Analysis:
The particle cannot be found in the regions where $V(x) > E$. This includes:
Regions $x $ where the potential energy is greater than $ E $.
For the given energy $E$, the particle cannot exist in the region where $V(x) > E$. directly.
The minimum total energy $E$ the particle must have is $ V_0$, which is the potential energy in the lowest region.
Physical Context:
This resembles a particle in a piecewise potential well, such as an electron in a stepwise potential well in semiconductor heterostructures.
(c) Inverted Step Potential
Potential Energy Function:
$ V(x) = 0 $ for $ x < 0 $
$ V(x) = V_0 $ for $ x = 0 $
$ V(x) = -V_1 $ for $a \leq x < b $
$ V(x) = 0 $ for $ x \geq b $
Analysis:
The particle cannot be found in the region $ V(x) > E $.
For $ E < 0 $, the particle is confined within the well.
The minimum total energy $E$ the particle must have is $-V_1$, as $ V(x) = -V_1 $ in the well region.
Physical Context:
This resembles a particle in an inverted potential well, such as an electron or positron trapped in a potential well with a finite depth.
(d) Symmetric Potential Well
Potential Energy Function:
$ V(x) = -V_1$ for $ -\frac{a}{2} \leq x \leq \frac{a}{2} $
$ V(x) > V_0 $ for $ |x| > \frac{a}{2} $
$V(x) $ increases symmetrically as $ |x| $ increases beyond $ \frac{a}{2}$
Analysis:
The particle cannot be found in the regions where $ V(x) > E$.
The minimum total energy (E) the particle must have is $-V_1 $, as $ V(x) = -V_1$ in the well region.
Physical Context:
This resembles a symmetric potential well, such as an electron in a quantum well or a shallow potential well in a crystal lattice.
Summary
The regions in which the particle cannot be found and the minimum total energy ( E ) for each case are summarized in the table below:
Case | Potential Energy Function Regions | Forbidden Regions | Minimum Energy ( E ) |
---|---|---|---|
(a) | ( x < a : V(x) = 0 ) | ( x \geq a ) | ( V_0 ) |
( x \geq a : V(x) = V_0 ) | |||
(b) | ( x < a : V(x) = V_0 ) | ( V(x) > E ) | ( V_0 ) |
( a \leq x < b : \text{linear increase} ) | |||
( b \leq x < d : V(x) = V_3 ) | |||
( x \geq d : V(x) = V_0 ) | |||
(c) | ( x < 0 : V(x) = 0 ) | ( x \leq 0, x \geq b ) | ( -V_1 ) |
( a \leq x < b : V(x) = -V_1 ) | |||
( x \geq b : V(x) = 0 ) | |||
(d) | ( -\frac{a}{2} \leq x \leq \frac{a}{2} : V(x) = -V_1 ) | ( V(x) > E ) | ( -V_1 ) |
( | x | > \frac{a}{2} : V(x) > V_0 ) |
The potential energy function for a particle executing linear simple harmonic motion is given by $V(x)=$ $k x^{2} / 2$, where $k$ is the force constant of the oscillator. For $k=0.5 \mathrm{~N} \mathrm{~m}^{-1}$, the graph of $V(x)$ versus $x$ is shown in Fig. 5.12. Show that a particle of total energy $1 \mathrm{~J}$ moving under this potential must 'turn back' when it reaches $x= \pm 2 \mathrm{~m}$.
Solving the equation:
$$ 1 = 0.25 x^2 $$
we get:
$$ x^2 = \frac{1}{0.25} $$
$$ x^2 = 4 $$
$$ x = \pm 2 $$
Therefore, a particle with total energy $1 \mathrm{~J}$ moving under the potential $V(x) = \frac{1}{2} k x^{2}$ must turn back when it reaches $ x < a : V(x) = 0 $ $x = \pm 2 \mathrm{~m} $ x < a : V(x) = 0 $.
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In Fig. 5.13 (i) the man walks $2 \mathrm{~m}$ carrying a mass of $15 \mathrm{~kg}$ on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of $15 \mathrm{~kg}$ hangs at its other end. In which case is the work done greater?
(a) The Casing of a Rocket in Flight Burns Up Due to Friction
The heat energy required for burning the rocket's casing is obtained at the expense of the rocket's kinetic energy. As the rocket moves through the atmosphere, air friction (drag force) acts on it, converting some of its kinetic energy into heat energy, which leads to the burning of the casing.
(b) Work Done by Gravitational Force on a Comet
The work done by gravitational force over one complete orbit of the comet is zero because gravitational force is a conservative force. In a closed orbit (complete orbit), the path starts and ends at the same point. The potential energy at the start and end points is the same, and thus, the net work done over one complete orbit is zero.
(c) Increase in Speed of an Artificial Satellite
As the satellite moves closer to Earth, it loses potential energy due to its decreasing altitude. However, since the total mechanical energy (the sum of kinetic and potential energy) decreases gradually, to conserve angular momentum, the satellite's speed increases. This increase in speed happens despite the gradual loss in the satellite's total mechanical energy.
(d) Work Done by the Man in Two Scenarios
Scenario (i): Man walking with 15 kg mass on his hands.
Here, the force applied by the man is vertical (to balance the weight), and displacement is horizontal.
Since work, ( W ), is given by $ W = \mathbf{F} \cdot \mathbf{d} \cos \theta $, and the angle $ x < a : V(x) = 0 \theta x < a : V(x) = 0 $ between force and displacement is $90^{\circ} $.
Therefore, work done, $W = Fd\cos(90^{\circ}) = F \cdot d \cdot 0 = 0$.
Scenario (ii): Man walking, pulling 15 kg mass via rope over pulley.
If we assume an ideal pulley with no friction and ignoring the weight of the rope, the force applied to pull the mass is vertically upward.
Since the man pulls the rope while walking horizontally, the displacement of the point where he applies the pulling force is still horizontal.
The angle $ \theta $ between the force (vertical) and displacement (horizontal) is $ 90^{\circ}$.
Therefore, work done here is also zero because $W = Fd\cos(90^{\circ}) = 0 $.
Conclusion
In both cases (i) and (ii), the work done by the man is zero because the force and displacement are perpendicular to each other.
Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Let's analyze each part carefully.
(a)
For a conservative force:
When the force does positive work on a body, mechanical energy is transferred to the body.
This work done reduces the potential energy of the body as it's converted to kinetic or other forms of energy.
Hence, decreases is the correct answer.
(b)
Work done against friction usually results in the conversion of kinetic energy into heat or other forms of energy.
Hence, there is always a loss of kinetic energy due to friction.
So, kinetic is the correct answer.
(c)
The rate of change of total momentum of a system (according to Newton’s Second Law) is proportional to the external force acting on the system.
Internal forces, as per Newton’s Third Law, cancel each other out in a closed system.
Thus, external force is the correct answer.
(d)
In an inelastic collision, total linear momentum and total energy are conserved, but total kinetic energy is not.
Therefore, the quantities which do not change are the total linear momentum and the total energy of the system of two bodies.
So, the correct answers are:
decreases
kinetic
external force
total linear momentum
Summary of Correct Alternatives
(a) decreases
(b) kinetic
(c) external force
(d) total linear momentum
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Here are the true or false statements along with reasons:
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
False.
Reason: In an elastic collision, the total momentum and total kinetic energy of the system (both bodies) is conserved. However, the momentum and energy of each individual body are not conserved separately. They can exchange momentum and energy during the collision.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
False.
Reason: The total energy of a system, including all forms of energy (kinetic, potential, thermal, etc.), is always conserved according to the law of conservation of energy. However, mechanical energy is conserved only if the forces doing work are conservative. In the presence of non-conservative forces (like friction), mechanical energy isn't conserved, although total energy remains constant by transforming into other forms like heat.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
False.
Reason: The work done over a closed loop is zero only for conservative forces (like gravitational force, spring force). For non-conservative forces (such as friction, air resistance), the work done is not zero over a closed loop because energy is dissipated as heat or other forms of non-recoverable energy.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
True.
Reason: In an inelastic collision, some of the initial kinetic energy is transformed into other forms of energy, such as heat, sound, or deformation energy. Therefore, the final kinetic energy is always less than the initial kinetic energy of the system.
Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
Answer:No, the total kinetic energy is not conserved during the short time of the collision while the balls are in contact. During this period, part of the kinetic energy is temporarily converted to potential energy due to deformation of the balls. The total kinetic energy is conserved only before and after the collision, not during the collision itself.
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
Answer:Yes, the total linear momentum is conserved during the short time of an elastic collision. According to Newton's Third Law, the forces between the two colliding balls are equal in magnitude and opposite in direction, ensuring that the total momentum remains constant throughout the collision process.
(c) What are the answers to (a) and (b) for an inelastic collision?
Answer:
For question (a) in an inelastic collision, the total kinetic energy is not conserved even before and after the collision. Part of the kinetic energy is transformed into other forms of energy like heat, sound, or potential energy due to deformation, and this energy is generally not fully recovered.
For question (b) in an inelastic collision, the total linear momentum is conserved during the short time of the collision, similar to the case of an elastic collision. This stems from Newton's Third Law, which ensures momentum conservation irrespective of the type of collision.
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic?
Answer:If the potential energy of two billiard balls depends only on the separation distance between their centres, this condition implies that the force during the collision is conservative. A collision involving conservative forces generally results in an elastic collision. Hence, the kinetic energy before and after the collision would be conserved, and there would be no permanent deformation or energy loss. Therefore, the collision is elastic.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to
(i) $t^{1 / 2}$
(ii) $t$
(iii) $t^{3 / 2}$
(iv) $t^{2}$
To determine how the power delivered to the body changes with time, we can use the known relations for motion under constant acceleration and the definition of power.
Steps to solve the problem:
Equations of Motion: For a body starting from rest under constant acceleration ( a ), $$ v = at $$where ( v ) is the velocity at time ( t ).
Kinetic Energy: The kinetic energy ( K ) at time ( t ) is given by, $$ K = \frac{1}{2} m v^2 $$ Substituting ( v = at ), $$ K = \frac{1}{2} m (at)^2 = \frac{1}{2} m a^2 t^2 $$
Power: Power ( P ) is the rate at which work is done, or the rate of change of kinetic energy, $$ P = \frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} m a^2 t^2 \right) $$ Simplifying, $$ P = m a^2 t $$
So, the power delivered to the body at time ( t ) is proportional to ( t ).
Conclusion
The correct option is: $$ \boxed{\text{(ii) } t}$$
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to
(i) $t^{1 / 2}$
(ii) $t$
(iii) $t^{3 / 2}$
(iv) $t^{2}$
To determine the relationship between the time $t$ and the displacement $s$, we start with the definition of power. Given that the power (P) is constant and power is defined as:
$$ P = F \cdot v $$
where (F) is the force and (v) is the velocity. For a body under constant power:
$$ P = \frac{dW}{dt} = F \cdot v $$
Firstly, let's calculate the kinetic energy (K):
$$ K = \frac{1}{2} m v^2 $$
Power is the rate of change of kinetic energy:
$$ P = \frac{d}{dt} \left(\frac{1}{2} m v^2\right) = m v \frac{dv}{dt} $$
Since ( F = m a ) and ( a = \frac{dv}{dt} ):
$$ P = m v \cdot a = m v \cdot \frac{dv}{dt} $$
Thus,
$$ P = \frac{d}{dt} \left(\frac{1}{2} m v^2\right) \Rightarrow P = v \cdot F $$
Now consider the constant power (P):
$$ P = m \cdot v \frac{dv}{dt} = m \cdot v a $$
Since ( P ) is constant:
$$ v \frac{dv}{dt} = \frac{P}{m} $$
Integrate both sides with respect to time (t):
$$ v \frac{dv}{dt} = \frac{P}{m} \Rightarrow v \frac{dv}{dt} \cdot dt = \frac{P}{m} \times dt $$
$$ \int{v ; dv} = \int{\frac{P}{m} ; dt} $$
$$ \frac{1}{2} v^2 = \frac{P}{m} t + C $$
Since ( v ) is initially zero when ( t = 0 ):
$$ C = 0 $$
Thus:
$$ v^2 = \frac{2P}{m} t $$
$$ v = \sqrt{\frac{2P}{m} t} $$
The displacement ( s ) is:
$$ s = \int{v ; dt} = \int{\sqrt{\frac{2P}{m} t} ; dt} $$
$$ s = \sqrt{\frac{2P}{m}} \int{t^{1/2} ; dt} $$
$$ s = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} $$
$$ s \propto t^{3/2} $$
Answer: The displacement in time ( t ) is proportional to $ t^{3/2} $, which matches option (iii).
A body constrained to move along the $z$-axis of a coordinate system is subject to a constant force $\mathbf{F}$ given by
$$
\mathbf{F}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \mathrm{N}
$$
where $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ are unit vectors along the $x$-, $y$ - and $z$-axis of the system respectively. What is the work done by this force in moving the body a distance of $4 \mathrm{~m}$ along the $z$-axis?
To calculate the work done by the force $\mathbf{F}$ in moving the body along the $z$-axis, we use the definition of work:
$$ W = \mathbf{F} \cdot \mathbf{d} $$
Here, $\mathbf{F}$ is the force vector and $\mathbf{d}$ is the displacement vector. Given:
$$ \mathbf{F} = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} , \text{N} $$
The displacement is along the $z$-axis, so:
$$ \mathbf{d} = 4 \hat{\mathbf{k}} , \text{m} $$
The work done is the dot product of $\mathbf{F}$ and $\mathbf{d}$:
$$ W = (\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) \cdot (4 \hat{\mathbf{k}}) $$
Using the properties of the dot product:
$$ \begin{aligned} W &= (\hat{\mathbf{i}} \cdot 4 \hat{\mathbf{k}}) + (2 \hat{\mathbf{j}} \cdot 4 \hat{\mathbf{k}}) + (3 \hat{\mathbf{k}} \cdot 4 \hat{\mathbf{k}}) \ &= 0 + 0 + (3 \cdot 4) \ &= 0 + 0 + 12 \ &= 12 , \text{J} \end{aligned} $$
Therefore, the work done is 12 joules.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10 \mathrm{keV}$, and the second with $100 \mathrm{keV}$. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass $=9.11 \times 10^{-31} \mathrm{~kg}$, proton mass $\left.=1.67 \times 10^{-27} \mathrm{~kg}, 1 \mathrm{eV}=1.60 \times 10^{-19} \mathrm{~J}\right)$.
Results:
Speed of the electron: $ v_e \approx 5.93 \times 10^7 ,\text{m/s} $
Speed of the proton: $ v_p \approx 4.38 \times 10^6 ,\text{m/s} $
Speed Comparison:
Clearly, the electron is faster than the proton.
Ratio of Their Speeds:
$$ \text{Ratio} = \frac{v_e}{v_p} = \frac{5.93 \times 10^7}{4.38 \times 10^6} \approx 13.54 $$
Conclusion: The electron is faster than the proton, and the ratio of their speeds is approximately 13.54.
A rain drop of radius $2 \mathrm{~mm}$ falls from a height of $500 \mathrm{~m}$ above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10 \mathrm{~m} \mathrm{~s}^{-1}$ ?
Work Done by Gravitational Force
First Half of the Journey:
The work done by gravitational force can be calculated using: $$ W_{g1} = m \cdot g \cdot d $$
Given:
$g = 9.81 , \mathrm{m/s^2} $
$ d = 250 , \mathrm{m} $ (half of the total journey)
The volume and mass calculations:
Volume, $ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2 \times 10^{-3} , \mathrm{m})^3 \approx 3.351 \times 10^{-5} , \mathrm{m^3} $
Density, $ \rho = 1000 , \mathrm{kg/m^3} $
Mass, $ m = \rho \cdot V \approx 3.351 \times 10^{-5} , \mathrm{kg} $
Therefore: $$ W_{g1} = 3.351 \times 10^{-5} , \mathrm{kg} \cdot 9.81 , \mathrm{m/s^2} \cdot 250 , \mathrm{m} \approx 0.00823 , \mathrm{J} $$
Second Half of the Journey:
Since the gravitational force acts equally in the second half: $$ W_{g2} = W_{g1} = 0.00823 , \mathrm{J} $$
Work Done by Resistive Force
The change in kinetic energy ( \Delta K ) is given by: $$ K_f = \frac{1}{2} m v_f^2 $$
Using the mass calculated earlier and final speed ( v_f = 10 , \mathrm{m/s} ): $$ K_f = \frac{1}{2} \cdot 3.351 \times 10^{-5} , \mathrm{kg} \cdot (10 , \mathrm{m/s})^2 \approx 1.675 \times 10^{-3} , \mathrm{J} $$
The total gravitational work ( W_g ) is: $$ W_g = W_{g1} + W_{g2} = 0.00823 , \mathrm{J} + 0.00823 , \mathrm{J} = 0.01646 , \mathrm{J} $$
Using the work-energy theorem to find the work done by resistive force ( W_r ): $$ W_r = K_f - W_g = 1.675 \times 10^{-3} , \mathrm{J} - 0.01646 , \mathrm{J} \approx -0.01479 , \mathrm{J} $$
Summary
Work done by gravitational force (first half): $ W_{g1} \approx 0.00823 , \mathrm{J} $
Work done by gravitational force (second half): ($ W_{g2} \approx 0.00823 , \mathrm{J}$
Work done by resistive force (entire journey): $W_r \approx -0.01479 , \mathrm{J} $
The negative sign for the resistive force indicates that it works against the motion of the raindrop.
A molecule in a gas container hits a horizontal wall with speed $200 \mathrm{~m} \mathrm{~s}^{-1}$ and angle $30^{\circ}$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
To evaluate the momentum conservation and the type of collision, let's analyze the scenario:
Momentum Conservation
Firstly, we'll decompose the velocity of the molecule into components.
Initial Velocity Components:
($ v_{x,i} = 200 \cos(30^\circ) $
$ v_{y,i} = 200 \sin(30^\circ) $
Final Velocity Components:
$v_{x,f} = 200 \cos(30^\circ) $ (unchanged direction of motion parallel to the wall)
$ v_{y,f} = -200 \sin(30^\circ) $ (reversed direction of motion perpendicular to the wall)
Where:
$\cos(30^\circ) = \frac{\sqrt{3}}{2} $
$ \sin(30^\circ) = \frac{1}{2} $
The components of velocities before and after the collision are:
$ v_{x,i} = 200 \cdot \frac{\sqrt{3}}{2} = 100\sqrt{3}$
$ v_{y,i} = 200 \cdot \frac{1}{2} = 100$
($ v_{x,f} = 100\sqrt{3} $
$ v_{y,f} = -100 $
We see that the x-component (parallel to the wall) remains the same, while the y-component (perpendicular to the wall) changes direction.
Momentum Conservation Analysis
Momentum parallel to the wall (x-direction): It is conserved.
Momentum perpendicular to the wall (y-direction): The momentum changes due to the change in direction.
The change in momentum perpendicular to the wall indicates an impulse from the wall, and so momentum is conserved in the system.
Type of Collision
To determine if the collision is elastic or inelastic, let's look at the kinetic energy:
Kinetic Energy before Collision:$$ K_i = \frac{1}{2} m v^2 = \frac{1}{2} m (200)^2 $$
Kinetic Energy after Collision:$$ K_f = \frac{1}{2} m v^2 = \frac{1}{2} m (200)^2 $$
Since the speed and, consequently, the kinetic energy are the same before and after the collision, the collision is elastic.
Conclusion
Momentum is conserved in the collision.
The collision is elastic because the kinetic energy is conserved.
A pump on the ground floor of a building can pump up water to fill a tank of volume $30 \mathrm{~m}^{3}$ in $15 \mathrm{~min}$. If the tank is $40 \mathrm{~m}$ above the ground, and the efficiency of the pump is $30 \%$, how much electric power is consumed by the pump ?
Work Done to Lift the Water:
Given:
Mass of the water $ m = 30000 , \mathrm{kg}$
Gravitational acceleration $ g = 9.8 , \mathrm{m/s}^2$
Height $ h = 40 , \mathrm{m} $
The work done ( W ) is: $$ W = mgh = 30000 , \mathrm{kg} \times 9.8 , \mathrm{m/s}^2 \times 40 , \mathrm{m} = 1.18 \times 10^7 , \mathrm{J} $$
Energy Consumed by the Pump:
Efficiency of the pump is 30%, so the energy consumed ( E ) is: $$ E = \frac{W}{0.3} = \frac{1.18 \times 10^7 , \mathrm{J}}{0.3} = 3.93 \times 10^7 , \mathrm{J} $$
Time Taken to Pump the Water:
The time ( t ) is: $$ t = 15 , \mathrm{min} = 900 , \mathrm{s} $$
Power Consumed:
Power ( P ) is: $$ P = \frac{E}{t} = \frac{3.93 \times 10^7 , \mathrm{J}}{900 , \mathrm{s}} = 4.36 \times 10^4 , \mathrm{W} $$
Converting to kilowatts: $$ P = 4.36 \times 10^4 , \mathrm{W} = 43.6 , \mathrm{kW} $$
Thus, the pump consumes 43.6 kilowatts of electric power.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed $V$. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision?
Fig. 5.14
For an elastic collision, both the momentum and kinetic energy are conserved. Let's analyze the given options:
Conservation of Momentum
The momentum before the collision is: $$ p_{\text{initial}} = mV + m \cdot 0 + m \cdot 0 = mV $$
The momentum after the collision should be equal: $$ p_{\text{final}} = mV_1 + mV_2 + mV_3$$
Conservation of Kinetic Energy
The kinetic energy before the collision is: $$ KE_{\text{initial}} = \frac{1}{2}mV^2$$
The kinetic energy after the collision should be: $$ KE_{\text{final}} = \frac{1}{2}mV_1^2 + \frac{1}{2}mV_2^2 + \frac{1}{2}mV_3^2 $$
Analysis of Options
Option (i)
Momentum: $ mV = m \cdot 0 + m \cdot \frac{V}{2} + m \cdot \frac{V}{2} = mV $
Kinetic Energy: $$ \frac{1}{2}mV^2 = \frac{1}{2}m \cdot 0^2 + \frac{1}{2}m \cdot \left(\frac{V}{2}\right)^2 + \frac{1}{2}m \cdot \left(\frac{V}{2}\right)^2 = \frac{1}{2}m \cdot \left(\frac{V^2}{4} + \frac{V^2}{4}\right) = \frac{1}{2}m \cdot \frac{V^2}{2} = \frac{1}{4}mV^2$$
This option fails to conserve kinetic energy as $\frac{1}{4}mV^2$ does not equal $\frac{1}{2}mV^2$.
Option (ii)
Momentum: $ mV = m \cdot 0 + m \cdot 0 + mV = mV $
Kinetic Energy: $$ \frac{1}{2}mV^2 = \frac{1}{2}m \cdot 0^2 + \frac{1}{2}m \cdot 0^2 + \frac{1}{2}mV^2 = \frac{1}{2}mV^2 $$
This option conserves both momentum and kinetic energy.
Option (iii)
Momentum: $mV = 3m \cdot \frac{V}{3} = mV$
Kinetic Energy: $$ \frac{1}{2}mV^2 = 3 \cdot \frac{1}{2}m \left(\frac{V}{3}\right)^2 = \frac{1}{2}m \cdot \frac{V^2}{3} \neq \frac{1}{2}mV^2$$
This option fails to conserve kinetic energy as $\frac{1}{6}mV^2 \neq \frac{1}{2}mV^2$.
Conclusion
Given that option (ii) conserves both momentum and kinetic energy, it is the only feasible outcome for an elastic collision.
Correct Answer: Option (ii)
The bob A of a pendulum released from $30^{\circ}$ to the vertical hits another bob $B$ of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
To determine how high bob A rises after the collision, we need to find certain quantities using the physics of pendulums and elastic collisions. Here’s the step-by-step approach:
Calculate Initial Height of Bob A:Since bob A is released from an angle of $30^\circ$, we can find the initial height using trigonometry.
Let ( L ) be the length of the pendulum.
The vertical height $ h_0$ from the lowest point is given by $ h_0 = L (1 - \cos \theta) $.
For (theta = 30^\circ $, $$ \cos 30^\circ = \frac{\sqrt{3}}{2} ] Therefore, [ h_0 = L \left(1 - \frac{\sqrt{3}}{2}\right) $
Initial Potential Energy of Bob A:The initial potential energy at height $ h_0$ is converted completely into kinetic energy just before the collision. $$ U_0 = m g h_0 $$
Kinetic Energy Before Collision:Since the tension does no work, and assuming an ideal pendulum: $$ K = U_0 = \frac{1}{2} m v^2 $$ So, we calculate the speed ( v ) using the conversion of potential energy: $$ m g h_0 = \frac{1}{2} m v^2 $$ $$ v = \sqrt{2 g h_0} $$
Elastic Collision:Since the collision is elastic, momentum and kinetic energy are conserved. After the collision, bob B will move, and bob A will come to rest momentarily, transferring all its velocity to bob B.
Since both bobs have the same mass and the collision is elastic: $$ v_{B_{\text{after}}} = v_A = \sqrt{2 g h_0} $$ Post collision, ( v_A \to 0 ), and bob B has the same velocity as bob A had before the impact.
Height Bob A Rises After Collision:After the collision, bob A has no kinetic energy and hence no height to rise due to the conservation of momentum and kinetic energy in a perfectly elastic collision of identical masses.
By considering all the steps and the principles involved, bob A will not rise after the collision because it will have zero velocity post the perfectly elastic collision. All its energy and momentum have been transferred to bob B, which will move with the speed $ \sqrt{2 g h_0}$ obtained just before the collision.
So, the height to which bob A rises after the collision is 0.
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5 \mathrm{~m}$, what is the speed with which the bob arrives at the lowermost point, given that it dissipated $5 \%$ of its initial energy against air resistance?
The speed with which the bob arrives at the lowermost point, given it dissipated 5% of its initial energy against air resistance, is approximately 5.28 m/s.
$$ v \approx 5.28 , \mathrm{m/s} $$
A trolley of mass $300 \mathrm{~kg}$ carrying a sandbag of $25 \mathrm{~kg}$ is moving uniformly with a speed of $27 \mathrm{~km} / \mathrm{h}$ on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of $0.05 \mathrm{~kg} \mathrm{~s}^{-1}$. What is the speed of the trolley after the entire sand bag is empty?
The speed of the trolley after the sandbag is completely empty is $ 8.125 , \text{m} , \text{s}^{-1}$.
Thus, the speed of the trolley after the entire sandbag is empty is $8.125 , \text{m/s} $.
A body of mass $0.5 \mathrm{~kg}$ travels in a straight line with velocity $v=a x^{3 / 2}$ where $a=5 \mathrm{~m}^{-1 / 2} \mathrm{~s}^{-1}$. What is the work done by the net force during its displacement from $x=0$ to $x=2 \mathrm{~m}$ ?
To find the work done by the net force during the displacement from $ x = 0 $ to $ x = 2 \text{ m} $), we can use the Work-Energy Theorem:
$$ W = \Delta K = K_f - K_i $$
The kinetic energy ( K ) is given by:
$$ K = \frac{1}{2} m v^2 $$
Given: $$ v = a x^{3/2}$$ $$a = 5 , \text{m}^{-1/2} , \text{s}^{-1}$$ $$ m = 0.5 , \text{kg} $$
Let's calculate the velocities at $ x = 0 $ and $ x = 2 , \text{m} $:
Initial velocity $v_i $ at $ x = 0 $:$$v_i = a \cdot (0)^{3/2} = 0 , \text{m/s} $$
Final velocity ( v_f ) at ( x = 2 , \text{m} ):$$v_f = a \cdot (2)^{3/2} $$ [ v_f = 5 \cdot (2)^{3/2} ]
Next, we calculate $ v_f $:
$$ (2)^{3/2} = 2^{1.5} = 2 \sqrt{2} \approx 2 \times 1.414 = 2.828 , \text{m/s} $$
$$v_f = 5 \times 2.828 = 14.14 , \text{m/s} $$
Now, compute the change in kinetic energy:
Initial kinetic energy ( K_i ):$$ K_i = \frac{1}{2} \cdot 0.5 \cdot (0)^2 = 0 , \text{J} $$
Final kinetic energy ( K_f ):$$ K_f = \frac{1}{2} \cdot 0.5 \cdot (14.14)^2 $$ $$ K_f = 0.25 \cdot 200 = 50 , \text{J}$$
Thus, the work ( W ) done by the net force is:
$$ W = K_f - K_i$$ $$ W = 50 , \text{J} - 0 , \text{J}$$ $$ W = 50 , \text{J} $$
The work done by the net force during the displacement from $ x = 0 $ to $ x = 2 , \text{m} $ is $ 50 , \text{J} $.
The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity $v$ perpendicular to the circle, what is the mass of the air passing through it in time $t$ ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts $25 \%$ of the wind's energy into electrical energy, and that $A=30 \mathrm{~m}^{2}, v=36$ $\mathrm{km} / \mathrm{h}$ and the density of air is $1.2 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the electrical power produced?
Using the results from the calculations:
(a) Mass of Air Passing Through the Circle
The mass of air passing through the area in time ( t ): $$ m = \rho \cdot A \cdot v \cdot t $$ Substituting the values ( \rho = 1.2 , \text{kg/m}^3 ), ( A = 30 , \text{m}^2 ), and ( v = 10 , \text{m/s} ) (converted from 36 km/h): $$ m = 1.2 \times 30 \times 10 \times t = 360 t , \text{kg} $$
(b) Kinetic Energy of the Air
The kinetic energy of the air passing through the area: $$ KE = \frac{1}{2} \rho \cdot A \cdot v^3 \cdot t $$ Substituting the values: $$ KE = \frac{1}{2} \times 1.2 \times 30 \times (10)^3 \times t = 1800 t , \text{J} $$
(c) Electrical Power Produced
The windmill converts ( 25% ) of the wind's kinetic energy into electrical energy: $$ P = 0.25 \cdot \frac{KE}{t} = 0.25 \times \frac{1}{2} \rho \cdot A \cdot v^3 $$ Substituting the values (\rho = 1.2 , \text{kg/m}^3), (A = 30 , \text{m}^2), and (v = 10 , \text{m/s}): $$ P = 0.25 \times \frac{1}{2} \times 1.2 \times 30 \times (10)^3 $$ The value calculated using these parameters is: $$ P = 4500 , \text{W} = 4.5 , \text{kW} $$
So, the electrical power produced is 4.5 kW.
A person trying to lose weight (dieter) lifts a $10 \mathrm{~kg}$ mass, one thousand times, to a height of $0.5 \mathrm{~m}$ each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies $3.8 \times 10^{7} \mathrm{~J}$ of energy per kilogram which is converted to mechanical energy with a $20 \%$ efficiency rate. How much fat will the dieter use up?
Solution:
(a) Work done against the gravitational force:
The work done by lifting the mass is given by: $$ W = m \cdot g \cdot h \cdot n $$
Substituting the values:
$m = 10 , \text{kg}$
$g = 9.8 , \text{m/s}^2$
$h = 0.5 , \text{m}$
$n = 1000$
$$ W = 10 , \text{kg} \cdot 9.8 , \text{m/s}^2 \cdot 0.5 , \text{m} \cdot 1000 $$
The work done is: $$ W = 49000 , \text{J} $$
(b) Fat used up:
Given:
Efficiency rate = 20%, so the effective energy conversion is: $$ 3.8 \times 10^7 , \text{J/kg} \cdot 0.20 = 7.6 \times 10^6 , \text{J/kg} $$
The amount of fat burnt can be calculated by: $$ \frac{49000 , \text{J}}{7.6 \times 10^6 , \text{J/kg}} $$
The amount of fat used up is: $$ \text{Fat} = 0.2 , \text{kg} \text{ (or 200 grams)} $$
Thus, the dieter will use up 200 grams of fat.
A family uses $8 \mathrm{~kW}$ of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of $200 \mathrm{~W}$ per square meter. If $20 \%$ of this energy can be converted to useful electrical energy, how large an area is needed to supply $8 \mathrm{~kW}$ ? (b) Compare this area to that of the roof of a typical house.
(a) Area Needed for Solar Panels
To find the area required to supply $8 \mathrm{~kW}$ of power when $20 %$ of direct solar energy can be converted to useful electrical energy, we use the given information:
Power needed: $8 \mathrm{~kW}$
Solar energy conversion efficiency: (20% = 0.2)
Incident solar power: $200 \mathrm{~W/m^2}$
The area needed can be calculated using the formula: $$ A = \frac{\text{Power needed}}{\text{Conversion efficiency} \times \text{Incident solar power}} $$
Substituting the given values: $$ A = \frac{8 \mathrm{~kW}}{0.2 \times 200 \mathrm{~W/m^2}} $$ $$ A = \frac{8000 \mathrm{~W}}{40 \mathrm{~W/m^2}} $$ $$ A = 200 \mathrm{~m^2} $$
So, 200 square meters of solar panel area is required to supply $8 \mathrm{~kW}$ of power.
(b) Comparison with Typical House Roof
Let's compare this required area with the roof area of a typical house.
Assuming the average roof area of a typical house is approximately 100 square meters (this value can vary, but it's a reasonable estimate for many residential homes).
Thus, the required solar panel area $200 ( \mathrm{m^2} )$ is twice the typical roof area of a house.
Summary
To supply $8 \mathrm{~kW}$ of power with an efficiency of $20%$ and an incident solar power rate of $200 \mathrm{~W/m^2}$, 200 square meters of area is needed.
This area is approximately two times the roof area of a typical house.
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Ask Chatterbot AINotes - Work, Energy, and Power | Class 11 NCERT | Physics
Comprehensive Notes on Work, Energy, and Power for Class 11
Introduction to Work, Energy, and Power
Everyday Language vs Physics Definitions
In everyday speech, terms like "work", "energy", and "power" are used frequently but often differently from their scientific meanings. For instance, a farmer ploughing a field, a student studying, and a runner exercising are all considered to be working. However, in physics, "work" holds a precise and specific definition. Similarly, "energy" and "power" are also well-defined in the realm of physics. This article aims to delve into these concepts, making them easier to understand for Class 11 physics students.
Importance in Physics
Understanding work, energy, and power is crucial as they are foundational concepts in physics. They facilitate comprehension of various physical phenomena, from everyday tasks to complex scientific processes.
The Scalar Product
Definition and Mathematical Representation
In physics, many quantities like displacement, velocity, and force are vectors—they have both magnitude and direction. The scalar product (or dot product) of two vectors is a way to multiply them to get a scalar (a quantity with magnitude only). The scalar product of vectors $\mathbf{A}$ and $\mathbf{B}$is defined as:$$ \mathbf{A} \cdot \mathbf{B} = AB \cos \theta $$ where $\theta$ is the angle between the vectors.
Properties of Scalar Products
Commutative Law: $\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$
Distributive Law: $\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C}$
Examples and Applications
For vectors (\mathbf{A}) and (\mathbf{B}), $$\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}} $$ $$ \mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}}$$ Their scalar product is given by: $$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z $$
Work and Kinetic Energy
Constant Forces
Work done by a constant force $\mathbf{F}$ when it moves an object through a displacement $\mathbf{d}$ is: $$ W = \mathbf{F} \cdot \mathbf{d} = Fd \cos \theta $$ This measure is crucial for understanding energy transfer in physical processes.
Variable Forces
For a variable force, the work done can be calculated by integrating the force over the path of the displacement: $$ W = \int_{x_i}^{x_f} F(x) , dx$$
Calculation of Kinetic Energy
Kinetic energy $(K)$ of an object with mass (m) and velocity (v) is: $$K = \frac{1}{2} mv^2 $$
Examples and Problems
Consider a cyclist who stops over a distance of 10m with a friction force of 200N: $$ W = Fd \cos \theta = 200 \times 10 \times \cos 180^\circ = -2000 , J $$
The Work-Energy Theorem
For Constant Forces
The work-energy theorem states that the work done by all forces on a body results in a change in kinetic energy: $$ K_f - K_i = W $$
For Variable Forces
For a variable force, $$ \Delta K = \int_{x_i}^{x_f} F(x) , dx $$
Mathematical Formulation and Examples
A moving block encountering a rough patch can have its kinetic energy calculated through the change in work done, considering force changes along the path.
Concept of Potential Energy
Gravitational Potential Energy
Potential energy due to gravity for a height (h): $$ V(h) = mgh $$
Potential Energy in Springs
For a spring with spring constant (k) compressed or extended by (x): $$ V(x) = \frac{1}{2}kx^2 $$
Conservative vs Non-Conservative Forces
Conservative forces have path-independent work done, and potential energy is defined. Non-conservative forces like friction depend on the path.
Examples and Problems
Calculate the gravitational potential energy of a 10kg mass lifted 2m: $$ V = mgh = 10 \times 9.8 \times 2 = 196 , J $$
Conservation of Mechanical Energy
Principle and Proof
Total mechanical energy $(E = K + V)$ of a closed system remains constant if only conservative forces act on it: $$ K_i + V(x_i) = K_f + V(x_f) $$
Examples and Applications
A ball dropped from height (H) converts potential energy at the top into kinetic energy at the bottom.
graph TD
A[Gravitational Potential Energy: mgh] --> B[Kinetic Energy: 1/2 mv^2]
Power
Definition and Calculation of Power
Power is the rate of doing work: $$ P = \frac{W}{t} $$
Average vs Instantaneous Power
Average Power: $ P_{avg} = \frac{W}{t} $
Instantaneous Power: $ P = \mathbf{F} \cdot \mathbf{v} $
Units of Power and Examples
SI Unit: Watt (W), where $1 , W = 1 , J/s $
Real-life Application: Calculating power usage in electrical devices like a 100W bulb running for 10 hours uses: $$ 100 \text{W} \times 10 \text{hrs} = 1 \text{kWh} = 3.6\times10^6 \text{J} $$
graph TD
A["Work (W)"] --> B["Time (t)"]
B --> C["Power (P)"]
Collisions
Elastic and Inelastic Collisions
Elastic Collisions: Both momentum and kinetic energy are conserved.
Inelastic Collisions: Only momentum is conserved; kinetic energy is not.
Momentum Conservation in Collisions
In all collisions, the total momentum is conserved: $$ \Delta \mathbf{p}_1 + \Delta \mathbf{p}_2 = 0 $$
Examples and Applications
Consider two colliding billiard balls. Their final velocities can be calculated using conservation laws.
Summary and Key Points
Key Formulas and Definitions
Work: $ W = \mathbf{F} \cdot \mathbf{d} $
Kinetic Energy: $K = \frac{1}{2} mv^2 $
Potential Energy: $V(x) = \frac{1}{2} k x^2$ for springs, (V(h) = mgh) for gravity.
Power: $ P = \frac{W}{t} $
Important Concepts to Remember
Difference between conservative and non-conservative forces.
Work-energy theorem and its applicability.
Conservation of mechanical energy.
Points to Ponder
Misconceptions and Clarifications
Work Done Definition: Always refer to the specific force doing the work.
Work as Scalar: Unlike mass and kinetic energy, work can be negative.
Mutual Forces Work Sum: The sum of work by mutual forces may not cancel out.
With these comprehensive notes, Class 11 students should now have a better grasp of the concepts of work, energy, and power, and be well-prepared for their physics studies.
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