Question

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2. A) Use the work-energy theorem to find the distance over which the force acts.B) If twice the force is exerted over twice the distance, how does the resulting increase in kinetic energy compare with the original increase in kinetic energy?

Answer 1

Answer:

A)$d=\dfrac{1}{2F}mv^2$

B)$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force  = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE'                          ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.