Question

A block of mass m slides down a smooth surface of a bowl of radius r from its rim to the bottom. The kinetic energy of the block at the bottom will be

Answer 1

Answer:

Answer will be MgR

Hope its correct :-)

Answer 2

Answer:

The K.E. of the block at the underside are $mgR.$

Explanation:

When a block slides from a particular height it'll reach the bottom with certain velocity and it's thanks to the law of conservation of energy. Total initial P.E. a block has are converted into the mechanical energy by the time it reaches the bottom of the bowl.

The given situation is represented within the below diagram.

Initially, velocity is zero. apart from attractive force, no other force is working on the block. So if the block slides down, its P.E. will decrease. The decrease in mechanical energy will appear as a rise in K.E. at the underside. So because the K.E. is increased, its velocity will increase.

So, Loss in mechanical energy is$U=mgR$

As we all know that loss in P.E. is up to the rise in K.E.

$\begin{aligned}K.E.&=U\\ K.E.&=mgR\end$

Hence, the kinetic energy is $mgR.$

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