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

\(\frac{1}{2} mgy^2_{0}\)
\(\frac{1}{2} mgy^2\)
mg(y - y0)
mgy0
The correct answer is (D) : \(mgy_0\)
Loss is potential energy = gain in kinetic energy
– (mg(y – y0) – mgy) = KE – 0
⇒ KE = mgy0
Read More: Work and Energy