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QPage 47: Q13: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q13 Solution: A van of mass 1500 kg travelling at a speed of is stopped in 10 …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the given $54\,\text{km h}^{-1}$ directly into $\frac12mv^2$, treating it as $54\,\text{m s}^{-1}$; this inflates the kinetic energy by a factor of $(\frac{1000}{3600})^2$. Always convert the speed to metres per second first, then compute the kinetic energy in joules and only then convert to calories before dividing by the stopping time.

A van of mass 1500 kg travelling at a speed of 54 km h−154 \text{ km h}^{-1} is stopped in 10 s. Assuming that all the mechanical energy lost appears as thermal energy in the brake mechanism, find the average rate of production of thermal energy in cal s−1\text{cal s}^{-1}.

Step-by-Step Verified Solution

**Answer:** Average rate of production of thermal energy � 4000 cal/s. **Step-by-Step Explanation:** Given: Mass of van = 1500 kg; Speed = 54 km/h = 15 m/s; Stopping time = 10 s. Kinetic energy:
KE=12×1500×152=168750 JKE = \frac{1}{2} \times 1500 \times 15^2 = 168750 \text{ J}
Loss in energy = 168750 J = 168750/4.186 = 40310 cal. Average rate = 40310/10=403140310/10 = 4031 cal/s � 4000 cal/s.[citation:2]

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