Question

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 1/2 mv2

Answer 1

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 $\dfrac{9}{2}mv^2$.

Answer 2

W work
F force 
s distance

If F = constant:

W₁ = F·s

If you triple the force and the distance:

W₂ = 3F · 3s = 9 F·s = 9 W₁