Question

When an empty lift is moving down with an acceleration of $\frac{g}{4} , \text{m/s}^2$, the tension in the cable is 9000 N. When the lift is moving up with an acceleration of $\frac{g}{3} , \text{m/s}^2$, the tension in the cable is:

(A) $16,000 , \text{N}$

B $18,000 , \text{N}$

C $12,000 , \text{N}$

D $15,000 , \text{N}$

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Answer

To solve the problem, we'll use Newton's second law of motion to analyze the scenarios where the lift is moving down and up, respectively.

Given Information:

  • When the empty lift is moving down with an acceleration of $ \frac{g}{4} , \text{m/s}^2 $, the tension in the cable is $9000 , \text{N}$.

Scenario 1: Lift Moving Downward

When the lift is moving down with an acceleration of $\frac{g}{4}$, we have:

  • The weight of the lift acting downwards: $mg$
  • The tension in the cable acting upwards: $T$
  • The downward acceleration: $\frac{g}{4}$

Using Newton's second law: $$ mg - T = m \left(\frac{g}{4}\right) $$

Given tension $T = 9000 , \text{N}$: $$ mg - 9000 = \frac{mg}{4} $$

Rearranging the equation to isolate $mg$: $$ mg - \frac{mg}{4} = 9000 $$ $$ \frac{3mg}{4} = 9000 $$ $$ mg = \frac{9000 \times 4}{3} = 12000 , \text{N} $$

So, the weight of the lift $mg = 12000 , \text{N}$.

Scenario 2: Lift Moving Upward

Next, when the lift is moving upwards with an acceleration of $ \frac{g}{3} , \text{m/s}^2 $, we have:

  • The weight of the lift acting downwards: $mg = 12000 , \text{N}$
  • The tension in the cable acting upwards: $T$
  • The upward acceleration: $\frac{g}{3}$

Using Newton's second law: $$ T - mg = m \left( \frac{g}{3} \right) $$

Substituting the known value of $mg$: $$ T - 12000 = \frac{mg}{3} $$ $$ T - 12000 = \frac{12000}{3} $$ $$ T - 12000 = 4000 $$

Solving for $T$: $$ T = 12000 + 4000 = 16000 , \text{N} $$

Thus, the tension in the cable is $16000 , \text{N}$ when the lift is moving up with an acceleration of $\frac{g}{3}$.

Final Answer:

(A) $16000 , \text{N}$

This matches the given option (A).


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