When an empty lift is moving down with an acceleration of 9.8/4 m/s^2, the tension in the cable is 9000 N. When the lift is moving up with an acceleration of 9.8/3 m/s^2, the tension in the cable is: (A) 16,000 N B 18,000 N C 12,000 N D 15,000 N

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).