Answer to Question #102186 in Mechanics | Relativity for dgfg

Question

An archer pulls the bowstring back for a distance of 0.39 m before releasing an arrow. The bow and string act as a spring of spring constant 400.1 N/m. At what speed will a 42.3 gram arrow leave the bow?

Answer 1

From the conservation of energy:

$$0.5mv^2=0.5kx^2$$

$$v=\sqrt{\frac{k}{m}}x$$

$$v=\sqrt{\frac{400.1}{0.0423}}0.39=37.93\frac{m}{s}$$