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

a) we know, that `U_(x)` = frictional force/normal reaction.
Frictional force = `U_(k) xx` normal reaction
`=0.1 xx 2kg wt = 0.1 xx 2 xx 9.8N=1.96N`
net effective force = (7-1.96)N= 5.04 N
Acceleration = `5.04//2ms^(-2) = 2.52 ms^(-2)`
Distance , s`=1//2xx2.52 xx 10 xx 10=126`m
work done by applied force = `7 xx 126J = 882J`
b) work done by friction =`1.96 xx 126=-246.96` J
c) work done by net force = `5.04 xx 126=-246.96J`
d) Change in the kinetic energy of the body
= work done by the net force in 10 seconds. =635.04 J (this is in accordance with work-energy theorem).