Question

A body of mass 5kg is moving with a momentum of 10 kg-m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 seconds. The increase in its kinetic energy is

Answer 1

Given,

Mass, m = 5kg

Momentum, p = 10 kg m/s

Force, F. =O.2 N

Time, t. = 10s

Now, Increase in kinetic energy,KE =

1/2 mv^2 - 1/2 mu^2

=> 1/2 m( v^2 - u^2 )

Here ,

Initial velocity,u = ?

Final velocity ,v = ?

And, Acceleration ,a. =?

Therefore, we have to find u and v first,

p = mu

=> 10 = 5(u)

=> u = 2m/s

Now,

F = ma

=> a = F/m

=> a = 0.2/ 5 m/s^2

=> a = 0.04 m/s^2

According to 1st equation of motion,

v = u + at

v = 2 + 0.04(10)

v = 2 + 0.4

v = 2.04 m/s

Now,

KE = 1/2 m (v^2 - u^2)

=> KE = 1/2× 5 [(2.4)^2 - (2)^2]

=>. KE = 2.5× ( 5.76 - 4 )

=>. KE =. 2.5 × 1.76

=>. KE =. 4.4 J

Ans.- The increase in kinetic energy is 4.4 J ( J= Joules).