The correct option is
A mg(μl+h)The free body diagram of the body showing forces acting on the body is as shown below.
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/622869/original_New_Bitmap_Image.bmp)
Resolving the weight of the body
mg along the surface and perpendicular to the surface as shown below.
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/622873/original_New_Bitmap_Image.bmp)
For the body to be in equilibrium, force can be equated as,
F=mgsinθ+μN(1)
N=mgcosθ(2)
Substituting the value of
N in equation (1), we get
F=mgsinθ+μmgcosθ
F=mg(sinθ+μcosθ)(3)
Taking an element of length
ds on the trajectory and let
dW be the work done by force
F to move the body by displacement
ds. Then
dW=F.ds
Substituting
F from (3),
dW=(mg(sinθ+μcosθ)).ds
Since force is directed along the trajectory, therefore angle between
F and
ds will be
0∘. Therefore,
dW=(mg(sinθ+μcosθ))×ds
dW=mg(dssinθ+μdscosθ)(4)
The elemental length
ds can be resolved into two components i.e horizontal
(dl) and vertical
(dh) as shown below.
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/622888/original_New_Bitmap_Image.bmp)
Therefore,
dssinθ=dh
dscosθ=dl
Putting these values in (4),
dW=mg(dh+μdl)
Integrating the above equation,
∫dW=∫(mg(dh+μdl))
⇒W=mg∫dh+mgμ∫dl
Now
∫dh=h and
∫dl=l
Hence
W=mgh+mgμl
⇒W=mg(h+μl)
For detailed solution watch next video.