Given, m=2kg,u=0, m=0.1,t=10s
Applied force F=7N
Force due to friction f=µmg=0.1×2×9.8=1.96N
Net force under which body moves F′=F–f=7–1.96=5.04N
Therefore acceleration with which body moves
a=F′m=

5.04

2

=2.52ms−2
Therefore distance moved by the body in 10 seconds
S=ut+

1

2

at2 =0×10+

1

2

×2.52×102 =126m
(a) Work done by the applied force in 10 s
W=F×S=7×126=882J
(b) Work done by the friction force in 10 s
W=–f×S=–1.96×126 =–246.9J=–247J
(c) Work done by the net force on body in 10 s
W=F′×S=5.04×126=635J
(d) Initial kinetic energy
=

1

2

mu2=

1

2

m×(0)2=0
Final kinetic energy
=

1

2

mv2=

1

2

×2×(25.2)2=635J
Change in kinetic energy = Final K.E. – Initial K.E. = 635 – 0 = 635
This shows that change in kinetic energy of the body is equal to work done by net force on the body.