A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Comput the
a) work done by the applied force in 10s,
b) work done by friction in 10 s,
c) work done by the net force on the body 10s,
d) change in kinetic energy of the body in 10s. Interpret your result.

Explanation

`m= 2kg, u = 0, m = 0.1, t = 10 s, F = 7N`.
Force due to friction `f = mu mg `
`f = 0.1 xx 2 xx 9.8 = 1.96N`
Net force under which body moves ,
`F. = F - f = 7 - 1.96 = 5.04 N`
So, acceleration with which body moves, `a = (F.)/m = (5.04)/(2) = 2.52 ms^(-2)`
The distance moved by the body in 10 s is
`s = u t + 1/2 at^2 " or " s = 0 xx 10 + 1/2 xx 2.52 xx (10)^2 = 126m`
a) Work done by the applied force in 10 s
`W = F xx s = 7xx126 = 882 J`.
b) Work done by the friction force in 10 s is
`W = - f xx s = -1.96 xx 126 = -246 J = -247 J`
c) Work done by the net force on body in 10 s is
`W = F xx s = 5.04xx126 = 635 J`.
d) Initial kinetic energy = `1/2 mv^2 = 1/2 xx m (0)^2 = 0`
Final kinetic energy = `1/2 mv^2 = 1/2 xx 2 xx (25.2)^2 = 635 J`
change in kinetic energy = `635 -0 = 635 J `
So, the change in kinetic energy of body is equal to work done by not force on the body.