# Work and spring problem

1. Oct 28, 2007

### relskid

1. The problem statement, all variables and given/known data

1. a force of 5 pounds compresses a 15-inch spring a total of 4 inches. how much work is done in compressing the spring 7 inches?

2. how much work is done in compressing the spring in exercise 1 from a length of 10 inches to a length of 6 inches?

2. Relevant equations

hooke's law: F=kd

3. The attempt at a solution

1.

F(k)=kx
5=k(4)
k=5/4

$$W = \int^7_0 (\frac{5}{4}x)dx$$

$$\frac{5}{4}[\frac{1}{2}x^2]^7_0$$

$$=\frac{245}{8} in-lb$$

note: i know that the above is correct, but it is needed for the second question, which i am unsure of.

2.

$$W = \int\begin{array}{c} 10 \\ 6 \end{array} (\frac{5}{4}x)dx$$

$$\frac{5}{4}[\frac{1}{2}x^2]\begin{array}{c} 10 \\ 6 \end{array}$$

$$\frac{5}{8}[x^2]\begin{array}{c} 10 \\ 6 \end{array}$$

$$\frac{5}{8}[(10^2)-(6^2)]\begin{array}{c} 10 \\ 6 \end{array}$$

$$=40 in-lb$$

note: in the second problem, i couldn't figure out how to fix the intervals, but it is [6,10]. that is actually the part that i'm a little confused about. is that the right interval? do i use the same k from the first problem? thanks in advance to any who can help.

2. Oct 28, 2007

### G01

Your work looks fine. Since you are using the same spring in part 2 as you are in part 1, your spring constant, k, won't change. For part two, to find the work, you are essentially finding the change in potential energy stored in the spring between compressions of 10in. and 6in., and this is exactly what your equation in part 2 is equal to so your interval is correct. Good job!