A van of mass 1200 kg is travelling at a speed of 54 kmh−1. When brakes are applied it comes to rest in 10 s. Assume that all the mechanical energy converts to thermal energy during the braking mechanism. Find the average rate of production of thermal energy.
Given,
Mass = 1200 kg
Speed v=54 kmh−1
= 54×(518) = 15 ms−1.
∴ Total K.E = 12mv2
=12×1200×(15)2
= 135000 J
We know, 1 cal=4.184 J
Total K.E = 1350004.184=32265 cal
It is given that total mechanical energy is converted into thermal energy.
Thermal energy = 32265 cal.
Rate of production of thermal energy
= 3226610=3226.6 cal/s
∴ The average thermal energy≈ 3227 cal s−1