Question

A body of mass 1 Kg initially at rest is moved by a horizontal force of 0.5 N on a smooth friction less table. Calculate the work done by the force in 10 S and show that it is equal to the change in kinetic energy of the body.

Answer 1

0.5 = a
s = ut + 1/2at^2
s =( 1/2 ) (0.5) 100
s = 25
work done = 0.5 × 25 = 12.5
v^2 = u^2 + 2as
v^2 = 2 × 0.5 × 25 = 25
change in kinetic energy = 1/2 m v ^2
=1/2 × 1 × 25
= 12.5

Answer 2

0.5 = a

s = ut + 1/2at^2

s =( 1/2 ) (0.5) 100

s = 25

work done = 0.5 × 25 = 12.5

v^2 = u^2 + 2as

v^2 = 2 × 0.5 × 25 = 25

change in kinetic energy = 1/2 m v ^2

=1/2 × 1 × 25

= 12.5