A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
Change in kinetic energy of the body in 10s and interpret your results

Explanation

Here, m = 2kg , u =0, F = 7 N , `mu = 0.1 W = 2, t 10 s`
Acceleration produced by applied force ,
`a = - (F)/(m) = (7)/(2) = 3.5 m//s^(2)`
force of friction, f ` = mu R = mu m g = 0.1 xx 2 xx 9.8 = 1.96 N`
Refardation produced by friction
`a_(2) = - (F)/(m) = (-1.96)/(2) = 0.98 m//s^(2)`
Net acceleration with which body moves, a `= a_(1) + a_(2)`
`= 3.5 - 0.98 = 2.52 m//s^(2)`
Distance moved by the body in 10 second from `S = Ut + (1)/(2) at^(2)`
`= 0 + (1)/(2) xx 2.52 xx (10)^(2) = 126 m`
From v = u + at
`v = 0 + 2.52 xx 10 = 25.2 m^(s-1)`
Final K.E= `(1)/(2) mv^(2) = (1)/(2) xx 2 xx (25 .2)^(2)`
= 635 J
Initial K.E `= (1)/(2) m u^(2) = 0`
` :.` Change in K.E = 635 - 0 = 635 J
This shows that change in K.E of the body is equal to workdone by the net force on the body