An object of mass 20 ~kg . is displaced by x=5 t^2 ~m (here t is time) by the application of a force. Then the ratio of the work done in times 3 ~s and 5 ~s is

Step-by-step solution

( Given, displacement of block, x=5 t^2 As, velocity v= d x d t Velocity of block, v= d d t (5 t^2 ) =5 2 t=10 t ~m / s Now, velocity of block at t=5 ~s , v₂=10 5=50 ~m / s and velocity of block at t=3 ~s is v₁=10 3=30 ~m / s Also, velocity of block at t=0 ~s , v₀=10 0=0 ~m / s Now, by work- K.E theorem, work done by a force = Change in kinetic energy So work done by force in first 3 sec , W₁= 1 2 m v₁^2- 1 2 m v₀^2= 1 2 m (v₁^2-v₀^2 ) And work done by force in first 5 sec , W₂= 1 2 m v₂^2- 1 2 m v₀^2= 1 2 m (v₂^2-