AB is a quarter of smooth circular track of radius `R=6m`. A particle P of mass `0.5kg` moves along the track from A to B under the action of the following forces.

a. A force `F_1` directed always towards the point B, its magnitude is constant and is equal to `20N`.
b. A force `F_2` directed along the instantaneous tangent to the circular track, its magnitude is `(15-10S)N`, where S is the distance travelled in metre.
c. A horizontal force of magnitude `30N`.
Find the work done by forces mentioned in (a), (b) and (c)

Answer

a. `120sqrt2J`; b. `45p(1-p)`; c. `180J`

Explanation

a. Work done by force `F_1`,
`dW_1=vecF_1*dvecs=F_1dscosalpha=F_1Rd thetacosalpha`
`=F_1Rd thetacos(pi/4-theta/2)`
`W_1=int dW_1=F_1Runderset0overset(pi//2)intcos(pi/4-theta/2)d theta`
`=F_1R((sin(pi/4-theta/2))/(-1/2))_0^(pi/2)`
`=-2F_1R[sin(pi/4-pi/4)-sinpi/4]`
`=2F_1Rxxsi npi/4=2F_1Rxx1/sqrt2=sqrtF_1R`
`=sqrt2xx20xx6=120sqrt2N`

b. Work done by `F_2`,
`W_2=int vecF_2*dvecs`
`=underset0overset(piR//2)int(15-10s)ds`
`=[15s-(10s^2)/(2)]^(piR/2)`
`=15(piR)/(2)-(5xxpi^2xxR^2)/(4)`
`=(15xxpixx6)/(2)-(5xxpi^2xx6^2)/(4)`
`=45pi-45pi^2=45pi(1-pi)=-302.8J`
c. Work done by `F_3`
`W_3=30xxR=30xx6=180J`