Two bodies A and B are released such that A slides down along the inclined plane and B falls freely as shown in the figure. If time taken by A to reach the ground is twice the time taken by B to fall freely then show that inclined plane is frictionless.

Solution

Let h be the height,

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be the length of the inclined plane and μ be the coefficient of friction between the block and inclined plane.
Therefore,

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If t is the time taken by block B to reach the ground, then equation of motion of B is,

space space space straight h equals 1 half gt squared

...(1)
Since, block A slides down the rough inclined plane, therefore acceleration of block A is,
a = g ( sin 30^o -

straight mu space cos space 30 to the power of straight o right parenthesis

straight g over 2 left parenthesis 1 space minus space square root of 3 space mu right parenthesis space

Here, A takes twice the time taken by B.
Therefore, equation of motion of block A is,

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That is, the inclined plane is frictionless.

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