How Much Work Does an Archer Do When Drawing a Bow?

[Mr Davis 97]

Homework Statement

[/B]
An archer pulls a bowstring back 0.440 m, and the spring constant of the bow is 545 N/m. Find the amount of work the archer did to draw the bow.

Homework Equations

[/B]
F = ma
E_i = E_f

The Attempt at a Solution

[/B]
It seems as though there are two ways of doing this: using Newton's laws or using conservation of energy.

a) Newton's laws:

W = Fd

F = -kx = 240 N

d = 0.440 m

W = Fd = 106 J

b) Conservation of energy

E_i = E_f

W_apply = U_final

W_apply = (1/2)kx² = 53 J
As one can see, these are two different answers. In fact, the latter is exactly half of the former. What am I doing wrong? Is either answer even right?

 

[Doc Al]

a) Newton's laws:

W = Fd

F = -kx = 240 N

d = 0.440 m

W = Fd = 106 J

This is incorrect. This calculation assumes the force is constant, but it's not: it varies from 0 to the maximum value.

b) Conservation of energy

Ei = Ef

Wapply = Ufinal

Wapply = (1/2)kx2 = 53 J

This is correct.

As one can see, these are two different answers. In fact, the latter is exactly half of the former. What am I doing wrong? Is either answer even right?

If you find the work correctly, by integrating ##\int F dx## since F is not constant, you'll get the same answer as you did in (b).

(Since the force varies linearly, you can cheat a bit and use the average force times the distance. The average force is 1/2 of the max force, so that will also give you the correct answer.)

 

[sk1105]

Was just writing pretty much what Doc Al said. You'll also notice that in your first method, W = Fd = Fx = -kx^2 ...twice your second answer.