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, up to when the particle comes to rest, is

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$\frac{Eμcosα}{sinα + μcosα}$

$\frac{Ecosα}{sinα + μcosα}$

$\frac{E}{sinα + μcosα}$

$\frac{E}{g (sinα + μcosα)}$

answer is A.

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Detailed Solution

From work energy theorem

$- m g \sin α . s - μmg cosα . s = 0 - E$

$s = \frac{E}{mg (\sin α + μcos α)}$

$∴ W o rk done against friction, W_{f} = μmg \cos α . s = \frac{μEcosα}{sinα + μcosα}$

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