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QPages 62-64: Q3: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q3 Solution: A 100 kg block is started with a speed of on a long, rough belt …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently assume the block’s kinetic‑energy loss is the same in every inertial frame and set ΔKE=0 in the moving frame, overlooking that kinetic energy depends on the observer’s velocity. In the frame moving with the block, the block starts at rest and ends with speed 2 m s⁻¹, gaining 200 J of kinetic energy; the belt‑holding external force does –200 J of work to keep the belt’s speed constant, reconciling the internal‑energy increase with the frame‑dependent kinetic‑energy change.

A 100 kg block is started with a speed of 2.0 m s−12.0 \text{ m s}^{-1} on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. b) Consider the situation from a frame of reference moving at 2.0 m s−12.0 \text{ m s}^{-1} along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m s−12.0 \text{ m s}^{-1}. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt. c) Find the work done in this frame by the external force holding the belt.

Step-by-Step Verified Solution

**Answer:** (a) ΔU=200\Delta U = 200 J (b) ΔKE=0\Delta KE = 0 in belt frame (c) Work done by external force = −200-200 J. **Step-by-Step Explanation:** Given: m=100m = 100 kg, u=2.0u = 2.0 m/s, μ=0.20\mu = 0.20. (a) Initial KE = 12×100×4=200\frac{1}{2} \times 100 \times 4 = 200 J. Since the block stops, this energy is dissipated as heat: ΔU=200\Delta U = 200 J. (b) In the belt frame, the block initially moves at −2.0-2.0 m/s and finally comes to rest relative to the belt. Change in KE = 0−12×100×4=−2000 - \frac{1}{2} \times 100 \times 4 = -200 J. (c) The external force holding the belt does work equal to the energy dissipated: W=−200W = -200 J.

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