Consider an object of mass 2 kg moved by an external force 20 N in a surface having coefficient of kinetic friction 0.9 to a distance 10 m. What is the work done by the external force and kinetic friction ? Comment on the result. (Assume g = 10 `ms^(-2)`)

Explanation

m = 2kg, d = 10 m, `F_(ext)=20N, mu_(k)=0.9`.
When an object is in motion on the horizontal surface, it experiences two forces.
(a) External force, `F_(ext)=20N`
(b) Kinetic friction,
`f_(k)=miu_(k)mg=0.9 times (2) times 10 = 18N`.
`" "`The work done by the external force
`W_(ext)=Fd=20 times 10=200J`
`" "`The work done by the force of kinetic friction `W_(k)=f_(k)d=(-18) times 10=-180J`. Here the negative sign implies that the force of kinetic friction is opposite to the direction of displacement.
The total work done on the object
`W_("total")=W_("ext")+W_(k)=200J-180J=20J`.
`" "` Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.