Question

A block of mass $m$ slides down a rough inclined plane of inclination $\theta$ with horizontal with zero initial velocity. The coefficient of friction between the block and the plane is $\mu$ with $\theta >ta{n}^{-1}\left(\mu \right)$. Rate of work done by the force of friction at time t is

• A. $\mu m{g}^{2}t\sin \theta$
• B. $m{g}^{2}t(\sin \theta -\mu \cos \theta )$
• C. $\mu m{g}^{2}t\cos \theta (\sin \theta -\mu \cos \theta )$
• D. $\mu m{g}^{2}t\cos \theta$

Answer 1

Answer: (C)

$\mu m{g}^{2}t\cos \theta (\sin \theta -\mu \cos \theta )$

Here, initial velocity , $u=0$

From the figure, net acceleration will be when it is sliding down,

frictional force=$\mu mgcos\theta$

$a=gsin\theta -ugcos\theta$........................(1)

We know, Power = work done /time= rate of work done..................(2)

Also, Power =force *velocity.............(3)

Now, from equation of motion,

$v=u+at$

$v=0+(gsin\theta -ugcos\theta )t$ (from eqn. (1)).....................(4)

From eqn. (2), (3) and (4),

work done /time= force *velocity,

=$\mu mgcos\theta .(gsin\theta -ugcos\theta )t$

=$\mu m{g}^{2}tcos\theta (sin\theta -\mu cos\theta )$