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Q.

A particle is moving with kinetic energy E, straight up an inclined plane with angle $α$ , the coefficient of friction being $μ$ . The work done against friction before the particle comes down to rest is:

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a

$\frac{Eμcos α}{\sin α + μcos α}$

b

$\frac{Ecos α}{\sin α + μcos α}$

c

$\frac{E}{g (\sin α + μcos α)}$

d

$\frac{E}{\sin α + μcos α}$

answer is A.

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Detailed Solution

By using work energy theorem

Question Image

$\begin{array}{l} - (μmgcos α) d - (mgsin α) d = 0 - E \\ \Rightarrow d = \frac{E}{mg (\sin α + μcos α)} \end{array}$
Work done against friction $= (μmgcos α) d$
$\begin{array}{l} = (μmgcos α) [\frac{E}{mg (\sin α + μcos α)}] \\ \frac{E μ \cos α}{\sin α + μ \cos α} \end{array}$

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