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Standard XII

Physics

Law of Conservation of Mechanical Energy

A block of ma...

Question

A block of mass 100 gm slides on a rough horizontal surface. If the speed of the block decreases from 10 m/s to 5 m/s, the thermal energy developed in the process is

3.75 J

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37.5 J

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0.375 J

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0.75 J

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Solution

The correct option is A 3.75 J
Thermal energy developed = Loss in kinetic energy $= \frac{1}{2} m (ν_{2}^{2} - ν_{1}^{2}) = \frac{1}{2} \times 0.1 \times (10^{2} - 5^{2}) = 3.75 J$

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A block of mass 100 gm slides on a rough horizontal surface. If the speed of the block decreases from 10 m/s to 5 m/s, the thermal energy developed in the process is

The correct option is A 3.75 J Thermal energy developed = Loss in kinetic energy =12m(ν22−ν21)=12×0.1×(102−52)=3.75J

The correct option is A 3.75 J
Thermal energy developed = Loss in kinetic energy $= \frac{1}{2} m (ν_{2}^{2} - ν_{1}^{2}) = \frac{1}{2} \times 0.1 \times (10^{2} - 5^{2}) = 3.75 J$