A block of mass ' $m$ ' with an initial kinetic energy ' $E$ moves up an inclined plane of inclination ' $\theta$ '. If ' $\mu$ ' is the coefficient of friction between the plane and the body, the work done against friction before coming to rest is

• A

  $\mu E \cos \theta$
• B

  $\frac{\mu E \cos \theta}{\sin \theta-\mu \cos \theta}$
• C

  $\frac{E \mu \cos \theta}{\cos \theta+\sin \theta}$
• Correct answer
  D

  $\frac{\mu E \cos \theta}{\sin \theta+\mu \cos \theta}$

Solution

Let before coming to rest body slides up to distance s . $\mathrm{h}=\mathrm{s} \sin \theta$
From the work- energy theorem, $\begin{aligned} & \Delta \mathrm{W}_{\hat{\varepsilon}}+\Delta \mathrm{W}_{\mathrm{mg}}=\Delta \mathrm{K} \\ & \Rightarrow-\mu \mathrm{mg}(\cos \theta) \mathrm{s}-\mathrm{mgs} \sin \theta=0-\mathrm{E} \\ & \Rightarrow \mu \mathrm{mgs} \cos \theta+\mathrm{mgs} \sin \theta=\mathrm{E} \\ & \Rightarrow \mathrm{~s}=\frac{\mathrm{E}}{\mathrm{mg}(\mu \cos \theta+\sin \theta)} \end{aligned}$ $\therefore$ Work one by friction, $\Delta W_{\mathrm{f}}=-\mu \mathrm{mgs} \cos \theta=\frac{\mu \mathrm{E} \cos \theta}{\mu \cos \theta+\sin \theta}=\frac{\mu \mathrm{E} \cos \theta}{\sin \theta+\mu \cos \theta}$

Hence, the correct answer is option: 'D'.