AB is a quarter of a smooth circular track of radius 4 m as shown in the figure. A particle P of mass 5 kg moves along the track from A to B under the action of the following forces

(1) A force `F_(1)` directed always towards point B, its magnitude is constant and equals 4 N,
(2) A force `F_(2)` that is directed along the instantaneous tangent to the circular track, its magnitude is (20-s) newton where s is the distance travelled in metre.
(3) A horizontal force F of magnitude 25 N.
Find the workdone by each force.

Explanation

(i) Work done by force `F_(1)` As shown in figure, let the particle be at point P at some instant of time t = t. The particle moves from position P to position Q in small interval of time dt. The direction of force on particle P will be in the direction PB. The small amount of work done in time dt is
`dW_(1) = F_(1).ds = F_(1) cos theta.ds = F_(1) cos theta.Rd theta`
`because ds = Rd theta = 16 cos theta d theta " "(because R = 4m and F = 4N)`
Now the total work done as the particle moves from A and B is given by
`W_(1) = int_(0)^(pi//4) 16 cos theta d theta = 16 [sin theta]_(0)^(pi//4) = 16 ((1)/(sqrt(2))) = (8 sqrt(2)) = 11.30` joule .......(1)
(ii) Work done by force `F_(2)` In the case, `d W_(2) = F_(2) xx ds = (20 - s)ds`
But `s = R theta and d s = Rd theta " "therefore d W_(2) = (20 - R theta) = (20 - 4 theta) xx 4 d theta`
`W_(2) = int_(0)^(pi//4) (80 d theta - 160 d theta) = 80 [theta]_(0)^(pi//4) - 16 [(theta^(2))/(2)]_(0)^(pi//4) = (80 xx pi)/(4) - (16)/(2) ((pi)/(4))^(2) = (62.83 - 4.93) = 57.9 J` ....... (2)
(iii) Work done by force `F_(3)`
The magnitude of `F_(3)` is 25 N which is always horizontal.
The net displacement of the particle is OB.
Hence the work done `W_(3) = 25 xx 4 = 100` joule ...........(3)