Question

A person is pulling a mass m from ground on a rough hemispherical fixed surface up to the top of the hemisphere with the help of a light inextensible string as shown in the figure. The radius of the hemisphere is $R$. Find the work done by the tension in the string. Assume that the mass is moved with a negligible constant velocity, $u$. (The coefficient of friction is $\mu$)

• A. $(R-1)Mg\mu$
• B. $(1+R)Mg\mu$
• C. $mgR(1-\mu )$
• D. $MgR(1+\mu )$

Answer 1

The correct option is D

$MgR(1+\mu )$

The net force acting on the sphere along the tangent (t) is given as

$T-f_k-mgsin\theta =ma\Rightarrow T-\mu N-mgsin\theta =ma$

Since the sphere moves slowly v is negligible and a = 0

$\Rightarrow T=\mu N+mgsin\theta$ .......(i)

Where $N-mgcos\theta =0$ since the sphere does not move radially(along the normal n).

And as u is negligible, and can assume there is no centripetal acceleration

$\Rightarrow N=mgcos\theta$

Using (i) and (ii), we obtain $T=mg(sin\theta +\mu cos\theta )$

Now the work done $dW$ shifting the ball through an infinitesimal distance $ds=dW=T.ds.cos\theta (\theta =0^{\circ }at$

every point during motion)

$\Rightarrow (Thetotalworkdoneinpullingthespherethroughadistancex=\frac{\pi R}{2})=W$

= $\underset{\frac{\pi R}{2}}{\overset{0}{\int }}T.ds\Rightarrow W=mg\underset{0}{\overset{\frac{\pi R}{2}}{\int }}(sin\theta +\mu cos\theta )ds$; putting $ds=Rd\theta$

We obtain $w=mgR\underset{00}{\overset{\frac{\pi }{2}}{\int }}(sin\theta +\mu cos\theta )d\theta =mgR(-Cos\theta +\mu Sin\theta {)}^{\frac{\pi }{2}}\Rightarrow W=mgR(1+\mu ).$