A body of mass 2 kg initially kg intially 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
(a) work done by the applied force in 10 s.
(b) work done by friction in 10 s .
work done by the net force on the body in 10 s.
(d) Change in kinetic energy of the body in 10 s and interpret your details .

Explanation

Frictional force ,
`f = muN = mu mg = 0.1 xx 2xx 9.8 = 1.96 N `
Net effective force ` F = f = 7 - 1.96 = 5.04 N `
Acceleration of body under the net force ,
`a = F/m =(5.04)/2 = 2.52 ms^(-2)`
Distance covered in 10 s , `S =0+1/2 at^(2) [ :. u = 0 ] `
` = 1/2 xx 2.52 xx10^(2)`
` = 126 m `
(a) work done by applied force ,
`W_(1)=(F)(s)`
` = 7 xx 126 = 882 J `
(b) Wrok done by friction ,
`W_(2) =-f(S)`
` = - 1.96 xx126 = - 246.96 = - 247 J `
(c ) Work done by net force = `5.04 xx126 = 635 J `
(d) Change in the kinetic energy of the body .
`DeltaK = DeltaW_(3)`
= 635 J
Interpretation : Work done by net force on the bod equal to the change in kinetic energy .