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
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) 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 results.

Answer 1

first we find displacement
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

Answer 2

First we find displacement

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