NEET-XII-Physics
25: Calorimetry
- #14A block of mass 100 g slides on a rough horizontal surface. If the speed of the block decreases from 10 m s-1 to 5 m s-1, find the thermal energy developed in the process.Ans : Given:
Mass of the block = 100 g = 0.1 kg
Initial speed of the block = 10 m/s
Final speed of the block = 5 m/s
Initial kinetic energy of the block = `` \frac{1}{2}\times 0.1\times {10}^{2}=5\,\mathrm{\,J\,}``
Final kinetic energy of the block = `` \frac{1}{2}\times 0.1\times {5}^{2}=1.25\,\mathrm{\,J\,}``
Change in kinetic energy of the block = 5 `` -`` 1.25 = 3.75 J
Thermal energy developed is equal to the change in kinetic energy of the block. Thus,
Thermal energy developed in the process = 3.75 J
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