Question

When a body of mass M slides down an inclined plane of inclination theta,through a distance s,the work done by surface is:(mu is friction coefficient) in terms of cos and sin.

Answer 1

Answer:

$W=Mg(sin\theta-cos\theta)\cdot s$

Explanation:

We are given that a body of mass M slides down an inclined plane of inclination theta .

Distance covered by body =s

Normal force on body =$Mgcos\theta$

Friction force =$\mu Mg cos\theta$

Applied force =ma=Ma

Net force =Applied force -friction force

$Net force =Mgsin\theta-Mgcos\theta$

Work done =$F\cdot S$

Substitute the values then we get

$W=Mg(sin\theta-cos\theta)\cdot s$

Hence, the work done by body=$Mg(sin\theta-cos\theta)\cdot s$

Answer 2

Answer:W = F.s cosô ( here cosô = 1 )

Here, F = friction force i.e. u×mg×cosØ

And s = displacement along the surface

Therefore,

W = u×mg×cosØ×s

Explanation: