A force of magnitude F_1 accelerates a body of mass m from rest to speed v. Another force F_2 accelerates a body of mass 2m from rest to a speed 2m. What is the ratio of the work done by F_2 to that of F_1?

Nov 24, 2016

${W}_{2} / {W}_{1} = 8$

Explanation:

We assume that the statement contains an error: where it says that the force ${F}_{2}$ accelerates a body of mass $2 m$ from rest to a speed $2 m$, we understand that it should say that ${F}_{2}$ accelerates said body to a speed $2 v$. In another case, the statement would not make much sense.

That said, we will solve the problem.

The work done by force on the body will be equal to the variation of the kinetic energy that it experiences:

$W = \Delta K = \frac{1}{2} m \left({v}^{2} - \cancel{{v}_{0}^{2}}\right) = \frac{1}{2} m {v}^{2}$

As both bodies start from rest, their initial velocities are two equal to zero.

The work done by the first force will be:

${W}_{1} = \frac{1}{2} m {v}^{2}$

and the one done by the second is:

${W}_{2} = \frac{1}{2} \left(2 m\right) \cdot {\left(2 v\right)}^{2} = 4 m {v}^{2}$

Then, the ratio between ${W}_{2}$ and ${W}_{1}$ is:

${W}_{2} / {W}_{1} = \frac{4 m {v}^{2}}{\frac{1}{2} m {v}^{2}} = 8$