Explanation :
When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by \(\dfrac{1}{2}mv^2\).
We know that work done is equal to the change in kinetic energy.
\(W=\Delta KE\)
Let kinetic energy increases from 0 to \(\dfrac{1}{2}mv^2\)
So, \(F.s=\Delta KE\)
\(F.s=\dfrac{1}{2}mv^2\)........................(1)
According to second condition if both force and distance is tripled, then kinetic energy increases by \(KE^'\).
So, \(3F\times 3s=\Delta KE^'\)
\(9F.s=\Delta KE'\)..........................(2)
Using equation (1) in (2)
\(\dfrac{9}{2}mv^2=\Delta KE^'\)
This is the resulting increase in kinetic energy.
Original increase in kinetic energy is \(\dfrac{9}{2}mv^2\).
