Question

A block of mass 30-0 kg is being brought down by a chain. If the block acquires a speed of 40-0 cm/s in dropping down 2-00 m, find the work done by the chain during the process.

Answer 1

The work done by the chain is -585.6 J.

Answer 2

Concept: Apply the Work-Energy Theorem: $W_{net} = \Delta K$. The net work is the sum of work done by gravity and work done by the chain.

Solution:

1. Convert units: $v_f = 40.0 \text{ cm/s} = 0.4 \text{ m/s}$.
2. Calculate initial and final kinetic energy: $K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(30.0 \text{ kg})(0 \text{ m/s})^2 = 0 \text{ J}$. $K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(30.0 \text{ kg})(0.4 \text{ m/s})^2 = \frac{1}{2}(30)(0.16) = 2.4 \text{ J}$.
3. Calculate work done by gravity: Gravity acts in the direction of displacement. $W_{gravity} = mgh = (30.0 \text{ kg})(9.8 \text{ m/s}^2)(2.00 \text{ m}) = 588 \text{ J}$.
4. Apply Work-Energy Theorem: $W_{net} = W_{gravity} + W_{chain} = K_f - K_i$. $588 \text{ J} + W_{chain} = 2.4 \text{ J} - 0 \text{ J}$. $W_{chain} = 2.4 \text{ J} - 588 \text{ J} = -585.6 \text{ J}$.

Explanation: The chain does negative work because its tension force acts upwards, opposite to the downward displacement of the block.