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


Part B
If triply the force is exerted over triply the distance, how does the resulting increase in kinetic energy compare with the original increase in kinetic energy?
Express your answer as an integer.

Answer 1

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

Answer 2

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