(a) A particle is acted upon by a force `vec(F)` given by `vec(F)=A cos omega t hat(i) + B hat(k)` and its position vector `vec(r)=a[(cos) ( omega t) hat(i)+ sin(omega t) hat(j)]+1/2 bt^(2)hat(k)`. Find the work done on the particle by the force `vec(F)` from time `t= pi // 2 omega ` to time `t=pi //omega`.
(b) A particle moves under a force `vec(F)= xy hat(i)+y^(2)hat(j)` and traverses along a path `y=4x+1`. Find the work done by the force when the particle is displaced from the point P(1, 5) to Q(2, 9).

Explanation

(a) The position vector of the particle is
`vec(r)=a[cos omega t hat(i)+sin omega t hat(j)]+1/2 bt^(2)hat(k)`
`:.` The velocity is given by, `vec(v)=(d vec(r))/(dt)=(-a omega sin omega t)hat(i)+(a omega cos omega t ) hat(i)+ ( bt) hat(k)`
Work done `W=int vec(F).vec(v)dt=-(a A omega)/(2) int_(pi//2 omega)^(pi//omega) sin 2 omega t dt + B b int_(pi//2omega)^(pi//omega)tdt`
`=(a Aomega)/(2)xx(cos 2 omega t)/(2 omega)|_(pi//2 omega)^(pi//omega)+Bb 1/2t^(2)|_(pi//2 omega)^(pi//omega)=(a A omega)/(2)[(1)/(2 omega)-(-1)/(2 omega)]+Bb. 1/2[(pi/omega)^(2)-((pi)/(2 omega))^(2)]`
`=(a A omega)/(2)xx1/omega+3/8 Bb ((pi^(2))/(omega^(2)))=1/2 aA+(3 pi^(2))/(8omega^(2))Bb`.
(b) Work done `W=vec(F).d vec(r)= int ( F_(x) hat(i)+F_(y)hat(j)).(dx hat(i)dy hat(j))`
`= int(F_(x)dx)+int(F_(y)dy)=int xydy + int y^(2)dy=int x ( 4x+1)dx+int y^(2)dy`
`=[4/3 x^(3)+(x^(2))/(2)]_(1)^(2)+[(y^(3))/(3)]_(5)^(9)=212` units.