Answer to Question #151031 in Physics for Tanveer Brar

Question

An archer uses her bow to shoot a 100g arrow. She pulls the bow back 58.7cm with a force of 250N. How fast will the arrow be travelling when it first leaves the bow?

Answer 1

If the arrow had experienced the force $F= 250N$ on the pass $s = 58.7cm = 0.587m$ then the work done over the arrow would be the following:

$$A = Fs = 146.75\space J$$

According to the work-energy theorem, the net work done by the forces on an object equals the change in its kinetic energy. If we assume, that the arrow did not change it's vertical position before leaving the bow, we can say, that the only force acted on the arrow was $F$. Thus:

$$A = \Delta K$$

where

$$\Delta K = \dfrac{mv^2}{2}-0$$

is the change in arrow's kinetic energy. Here $m = 100g = 0.1kg$ is its mass, and $v$ its speed after is left the bow. Obtain:

$$A = \dfrac{mv^2}{2}\\ v = \sqrt{\dfrac{2A}{m}} = \sqrt{\dfrac{2\cdot 146.75}{0.1}}\approx 54.2m/s$$

Answer. 54.2 m/s.