# Math Help - How much work is done...

1. ## How much work is done...

A force of 7 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 12 in. beyond its natural length?

Now I thought work was just force times distance. So 7 times 4...but my answer is wrong. What's the deal?

2. You must find the spring constant and integrate, not just multiply 7 times 4.

$F=kx$

$7=k(8)$

$k=\frac{7}{8}$

$\frac{7}{8}\int_{0}^{12}xdx$

3. Originally Posted by galactus
You must find the spring constant and integrate, not just multiply 7 times 4.

$F=kx$

$7=k(8)$

$k=\frac{7}{8}$

$\frac{7}{8}\int_{0}^{12}xdx$
But that's also incorrect. It'd be x^2/2 times 7/8, 12 squared is 144 divided by 2 is 72 times 7/8 is 63, and it's telling me 63 is incorrect...