S.O.S. Mathematics CyberBoard

A force of 10 lb is required to hold a spring stretch 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

The answer my book gives me is 15/4 ft-lb. Is there anyway to do this without converting to feet? The solution in my book converts everything to feet and then finds the answer. Is here any other way to solve this just with the info given?

I understand that you must use the formula
f(x) = kx (Hooke's Law)

Thanks.

You either have to use inches or feet, which requires a conversion.
It should not be a big problem, however, since 6 in = 1/2 ft and
4 in = 1/3 ft

determine the spring constant k...
F = kx
10 lbs = k*(1/3 ft)
30 lbs/ft = k

W = (1/2)k*x^2
W = (1/2)(30 lbs/ft)(1/2 ft)^2
W = 15/4 ft-lbs

If you wish to use inches...
F = kx
10 lbs = k(4 in)
(5/2) lbs/in = k

W = (1/2)k*x^2
W = (1/2)(5/2 lbs/in)(6 in)^2
W = 45 in-lbs
note that (45 in-lbs)(1 ft/12 in) = 45/12 ft-lbs = 15/4 ft-lbs