A.) How much work must be done in order to compress a spring with an elastic constant of 400 lb / in and 1.75 inches?

Question

*Note: 1.75" is the compression & 400#/in is the k constant.

B.) If 50 ft lbs of work is done on the same spring (in part a) in order to compress it, by how much is the spring shortened?

Answer 1

(a)
W=kx^2 (watch units)
(b)
use same equation, with W=50 ft-lbs = 600 ft-in.
solve for x.

Answer 2

Correction to equation
(a)
W=kx^2 ÷2 (watch units)
(b)
use same equation, with W=50 ft-lbs = 600 ft-in.
solve for x.

Answer 3

So "MathMate" you're saying that the equation I should use is the following:

A) 1/2 * 400 #/in * 1.75"^2 = 612.5 lbs

B) 1/2 * 600 #/in * 1.75"^2 = 918.75 lbs

There is no subtraction between the two? for part b, I need the length of the spring not lbs...