A body of mass 2 kg is rest on a smooth horizontal surface . When a horizontal force of `0.5` N acts on this bofy it is displaced in the direction of the force . Find the work done by the force in `8.0` s .Show that this work is equal to the change in kinetic energy of the body .

Explanation

Acceleration produced on a body of mass 2kg by force of 0.5 N .
`a = F/m`
` :. A = (0.5)/2 " " :. A = 0.25 ms^(-2)`
The displacement of the body in 8 s ,
`d = v_(0)t +1/2 at^(2)`
here, `v_(0)=0, a = 0.25 ms^(-2) , t = 8.0 s `
`d - 1/2 xx0.25 xx64`
` :. d = 8 m `
` :. ` Work W = fD = `0.5 xx8 = 4 J `
Alternative Method :
The velocity of body afer time 5 s from rest position ,
`v = v_(0)+at, v = 0 +F/m xxt`
` :. v = (0.5)/ 2xx 8 = 2 m//s `
` :. ` Change in kinetic energy of body .
`DeltaK =1/2 mv^(2) -1/2 mv_(0)^(2)`
` = 1/2 m(v^(2)-v_(0)^(2))=1/2 xx2x(4-0)`
` :. DeltaK = 4J `
` :. ` According to work energy theorem `DeltaK = W`
` :. ` Work =4 J .