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`. Calculate the
(a) work done by applied force in 10s. (b) work done by friction in 10s.
(c ) work done by the net force on the body in 10s.
(d) change in K.E. of body in 10s, and interpret your result.

Explanation

`m = 2 , F_(1) = 7 N , mu = 0.1 u = 0 `
Frictional force ` = F_(2) = mumg = 0.1 xx 2xx 9.8 `
` = 1.96 N `
Net acceeration ` a = (F_(1)-F_(2))/m = (7-1.96 )/2 `
` = 2.52 m//s^(2)`
Therefore , distance moved by the body in 10 s
` s = ut +1/2 xxat^(2)`
` = 0 +1/2 xx 2.53 (10)^(2)`
= 125 m
(a) `W_(1) =F_(1) s = 7 xx 126 = 882 J `
(b) `W_(2) ` = Work done by force of friction
` = - F_(2) xx s = - 1.96 xx126 = - 246 .96 J `
(c ) Work done by net force ` = W_(3) = (F_(1)-F_(2))s`
` = (7 -1.96) xx126`
`= 635 . 04 J `
(d) Intial K.E = 0
` v = u +at`
`= 0 +2.52 xx10`
` = 25.2 " m "s^(-1)`
Final K.E = `1/2 mv^(2)`
`= 1/2 xx 2 xx (25.2)^(2) = 635.04 J `
Change in K.E = `635.04 J - 0 `
` = 635.04 J `
This is equal to work done by the net force on the body .