# An archer pulls back on a 0.21 kg arrow 1.5 m using a bow. He then shoots the arrow and it leaves the bow traveling at 112 m/s. What was the elastic potential energy stored In the bow?

Nov 29, 2015

$1317.12 \text{J}$

#### Explanation:

The elastic potential energy of the bow is converted into the kinetic energy of the arrow.

$\text{KE} = \frac{1}{2} m {v}^{2}$

$= 0.5 \times 0.21 \times {112}^{2} = 1317.12 \text{J}$

So this was the elastic potential energy stored in the bow.