A block of mass M slides along the sides of a bowl as shown in the figure. The walls of the bowl are frictionless and the base has coefficient of friction 0.2. If the block is released from the top of the side, which is 1.5 m high, where will the block come to rest ? Given that the length of the base is 15 m

1 m from P

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Mid point

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2 m from P

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At Q

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Solution

The correct option is B
Mid point

Potential energy of block at starting point = Kinetic energy at point P = Work done against friction in traveling a distance s from point P.

$∴ m g h = f s = μ m g s \Rightarrow s = \frac{h}{μ} = \frac{1.5}{0.2} = 7.5 m$

i.e. block comes to rest at the mid point between P and Q.

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