Question

The work done in pulling a body of mass 5 kg along an inclined plane (angle 60º) with coefficient of friction 0.2 through 2m, will be

Answer 1

Answer: 104.47 J

The work done is given by:

W=F.s

Where, F is the net force acting on an object and s is the displacement.

On an inclined plane, the net force acting on the object is given by component of the weight acting along the inclined plane $(mg sin\theta)$and the force of friction $(\mu mg)$

$F=(mg sin\theta)+(\mu mg)$

Here, $\theta = 60^o$

mass,$m= 5kg$

coefficient of friction, $\mu=0.2$

displacement, $s=2 .0 m$

acceleration due to gravity, $g=9.8 m/s^2$

$F=(5kg\times 9.8 m/s^2 sin 60^o)+( 0.2\times 5 kg \times 9.8 m/s^2)=104.47 J$

Answer 2

Answer:

Force of friction is down the incline,

f=umgcos@, workdone by friction is Wf =umgcos@ds=-10 as well as one component of weight is acting incline downwards.

Wmg = mgsin@ds=-50sqrt3. Net workdone =change in kinetic energy. Wmg + Wf +W =0.  

W = -(Wmg + Wf). W = (10+50sqrt3) and W= 96.5 J