A particle of mass `'m'` moves along the quarter section of the circular path whose centre is at the origin . The radius of the circular path is `'a'`. A force `vec(F)=yhat(i)-xhat(j)` newton acts on the particle, where `x,y` denote the coordinates of position of the particle. Calculate the work done by this force in taking, the particle from point, `A(a,0)` to point `B(0,a)` along the circular path.

Answer

`-(pia^(2))/(2)`

Explanation

Work done by force `F ,`
`w=int vec(F).dvec(r)=int (yhat(i)-xhat(j)).(dxhat(i)+dyhat(j))`
`:. x^(2)+y^(2)=a^(2) :. xdx +y dy=0 `
`rArrW=int(y((-ydy)/(x))-xdy)`
`-int((x^(2)+y^(2)))/(x)dy`
`=-int_(0)^(a)(a^(2))/(sqrt(a^(2)-y^(2)))dy=-(pia^(2))/(2)`
Aternate Method
It can be observed that the force is tangent to the curve at each point and the magnitude is constant. The direction of force is opposite to the direction of motion of the particle.
`:. ` work done =(force) `xx` (distance)
`=-sqrt(x^(2)+y^(2))(pia)/(2)=-axx(pia)/(2)=-(pia^(2))/(2)J`
Ans. `w=-(pia^(2))/(2)J`