Figure. A smooth circular path of radius R on the horizontal plane which is quarter of a circle A block of mass m is taken from position A to B under the action of a constant force F. Calculate the work done by force F.
a. If it is always directed horizontally
b. If the block is pulled by a force F which is always tangential to the surface
c. Block is pulled with a constant force F which is always directed towards the point B

Explanation

a. `W_R=int vecF*dvecS=intFdscostheta`
or `W=underset0oversetRintFdx=FR`
As the block moves from A to B, the displacement of the block in the direction of force is equal to radius R.

Therefore, the work by the constant force `F` is `W=FR`.
b. If the block is pulled by a force F which is always tangential to the surface, in this case, force and displacement are always parallel to each other. The displacement of the block in the direction of force is `pi/2R`.
Thus, the work done by the force is
`W=F((piR)/(2))=pi/2FR`
c. Block is pulled with a constant force F which is always directed towards the point B. In this case, angle between force vector and displacement vector is varying.
In figure the angle between `vecF` and `dvecS` is `theta`. Block is at angle `alpha` from vertical The magnitude of `ds` is R. `dalpha` The relation between `theta` and `alpha` is

`(pi/4+alpha/2)+theta=pi/2`
`:. theta=pi/2-alpha/2`
Thus, `dW=vecFdvecs=Fdscostheta=F(Rdalpha)cos(pi/4-alpha/2)`
or `dW=(FR)/(sqrt2)(cos alpha/2+sinalpha/2)dalpha`
`=(FR)/(sqrt2)(underset0overset(pi//2)intcosalpha/2dalpha+underset0overset(pi//2)intsinalpha/2dalpha)`
or `W=FRsqrt2`