Work Energy Power

JEE Main level, single correct, medium

85 of 183

Question 85 of 183

In the given figure, the block of mass mm is dropped from the point ′A′'A'. The expression for kinetic energy of block when it reaches point ′B′'B' is

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Answer: (A) mgy0mg{y}_{0}

Here, gain in kinetic energy==loss in potential energy
Now loss in potential energy of particle in falling from height yy to y0{y}_{0} is mg(y−y0)mg\left(y-{y}_{0}\right).
So, ΔKE=\Delta KE=K−0=−[mg(y−y0)−mgy]K-0=-\left[mg\left(y-{y}_{0}\right)-mgy\right]
Thus, kinetic energy at point BB is K=mgy0K=mg{y}_{0}

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