Question

A body of mass 5kg has momentum of 10kg m/s .when a force of 0.2n is applied on its for 10sec, what is the change in its kinetic energy

Answer 1

First of all let me tell you an important fact:

If,

   % Increase in momentum (P) = n

Then,

  % Increase in kinetic energy (E) = n²+200n/100 → (1)

Initial Momentum = P = 10 N-s

Mass = 5 kg

Force = 0.2 N

Time for which force acted = 10 s

Now,

        Initial Kinetic Energy = E = ⅟₂mv² = ⅟₂(m)(p/m)²

                                                    = ⅟₂(5)(10/5)²

                                                    = 10 J

Now,

       ∆P = F*t

    P’- P = 2

         P’ = P+2 → (2)

From eq (2) we see that increase in momentum is (2/10)*100 = 20%

So from eq (1),

    Increase in Kinetic Energy = 44% of E    

                                                   = (44%)*10