JEE Physics 2023 Question with Solution

Given: mass of block is $1 \, \text{kg}$, average radius of tube is $15 \, \text{cm}$, so the vertical drop from top to bottom is $30 \, \text{cm} = 0.3 \, \text{m}$. Initial speed at the top is $22 \, \text{m/s}$ and the block finally stops at the bottom.

Find: the work done by the tube on the block.

Use the work-energy theorem:

$$W_f + W_{\text{gravity}} = \Delta K$$

Substitute the known values:

$$W_f + 10 \times 0.3 = 0 - \frac{1}{2} \times 1 \times (22)^2$$ $$W_f + 3 = -242$$ $$W_f = -245 \, \text{J}$$

Therefore, the work done by the tube on the block is $-245 \, \text{J}$.

Given: the tube is rough, so the contact force from the tube does non-conservative work on the block. The block moves from the top to the bottom and comes to rest.

Find: net work done by the tube.

The change in kinetic energy is from

$$K_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times 22^2 = 242 \, \text{J}$$

to

$$K_f = 0$$

So,

$$\Delta K = K_f - K_i = -242 \, \text{J}$$

Gravity assists the downward motion through a height difference of $0.3 \, \text{m}$, so

$$W_{\text{gravity}} = mgh = 1 \times 10 \times 0.3 = 3 \, \text{J}$$

Now apply

$$W_{\text{tube}} + W_{\text{gravity}} = \Delta K$$ $$W_{\text{tube}} + 3 = -242$$ $$W_{\text{tube}} = -245 \, \text{J}$$

Hence, the tube does negative work of $245 \, \text{J}$, so the required answer is $-245$.