Question

An archer's bow can be treated as a spring with

k = 2,800 N/m.

If the bow is pulled back a distance 0.17 m before releasing the arrow, what is the work done on the bow as the bowstring is pulled back into position?
J

Answer 1

First, since the force to draw the bow is not constant, but is modeled as a spring with a spring-constant (k) of 425N/m, we can write an expression for how the force varies with the distance pulled back:

F(x)=kx.

Now, energy is force * distance, but since this force varies, we need to integrate:

E(x) =

Answer 2

k = 2,800 N/m.

x= 0.17 m

the work done is = 1/2 k x^2

=0.5 * 2800 * (0.17)^2 = 40.46 J

Answer 3

Potential energy stored in a spring is (1/2)*kx^2 where x is the amount of stretch.

This PE will be translated into the KE of the arrow.

K.E= 1/2*K*X^2=0.5X2800X0.17X0.17=40.46 JOULS