Question

Please help, assume that 10 ft lbs of work is required to stretch a spring 1 ft beyond its natural length. What is the spring constant?

Answer 1

My answer is 10 lb/ft but the answer in the book is 20lb/ft

Answer 2

w=fd
10ft lbs = f (1ft)
F=10lbs
F(x)=kx
10lbs=k(1ft)
k=10lbs/ft

Answer 3

Energy(spring) = 1/2 * k * x^2
Plugging in your numbers, you have:
10 = 1/2 * k * (1)^2
Solving for k, k = 20

Answer 4

The energy it takes to stretch a spring is \[\frac{kx^2}{2}\].
Therefore, \[E = \textrm{10 ft * lbs} = \frac{k * 1^2}{2}\]
Rearranging the equation gives k = 20 lb / ft