Question

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

Fig.

• A. $\frac{1}{2} mgy^2_{0}$
• B. $\frac{1}{2} mgy^2$
• C. mg(y - y0)
• D. mgy0

Answer 1

Answer: (D)

mgy0

Answer 2

The correct answer is (D) : $mgy_0$
Loss is potential energy = gain in kinetic energy
– (mg(y – y 0) – mgy) = KE – 0
⇒ KE = mgy 0