Question

A body of mass 2 kg initially 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 work done by the net force on the body in 10 s.

Answer 1

f = μmg

f = 0.1 X 2 X 10

f = 2 N

Net force = F - f

                = 7 - 2

                = 5 N

F = ma

5 = 2a

a = 2.5 m/s²

s = ut + 1/2 at²

s= 0 X 10 + 1/2 X 2.5 X 10²

s= 1/2 X 2.5 X 100

s= 250/2

s= 125 m

W = Fs

W = 5 X 125

W = 625 Joules

Answer 2

Fnet=F-uN where u is coefficient of friction

ma=7N-(0.1)2 x 10 N

=7N-2N=5N

a =5/2m/sec^2

now use kinematic formula

s=ut+1/2at^2

=0+1/2 (5/2)(10)^2=125m

a) workdone by applied force =force x displacement =7N x 125m=875 Nm

b) workdone by frictional = -f x S

= -2 x 125=-250Nm

c) workdone by net force =(2x 5/2N)x 125=625Nm

d) apply conservation of energy theorem

workdone by net force=change in kinetic energy =625Nm