A body of mass of 3 kg initially at rest wakes under the action of an applied horizontally force of 10 N on a table with co-efficient of kinetic friction = 0.3, then what is the workdone by the applied force in 10 s :

Explanation

Applied force = 10 N
Opposing friction force `f=M_(k).N=M_(k)mg`.
`=0.3xx3xx9.8=8.82 N`.
Net accelerating force `F-f=10 N - 8.82 N`
`=1.18 N`
Acceleration `a = ("force")/("mass")=(8.82N)/(3kg)=2.94 ms^(-2)`
Distance covered in 10s (assuming w = 0)
`s=0+(1)/(2)at^(2)=(1)/(2)xx2.94xx(10^(2))=147 m`
there force workdone by a applied force,
`W = Fs = 10 xx 147`
`W=1470 J`