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==2.52ms−2Therefore distance moved by the body in 10 seconds
S=ut+at2 =0×10+×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
=mu2=m×(0)2=0Final kinetic energy
=mv2=×2×(25.2)2=635JChange 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.