JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 14)
In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is
$${1 \over 2}mg\,{y_0}^2$$
$${1 \over 2}mg\,{y^2}$$
$$mg(y - {y_0})$$
$$mg{y_0}$$
Explanation
Loss in potential energy = gain in kinetic energy
$$-$$ (mg(y $$-$$ y0) $$-$$ mgy) = KE $$-$$ 0
$$\Rightarrow$$ KE = mgy0
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