work force
posted by david .
a force of 10 pounds is required to stretch a spring of 4 inches beyond it's natural length. assuming hooke's law applies, how much work is done in stretching the spring from its natural length to 6 inches beyond its natural length.
what i did so far:
F(x)=kx
10=k4
k=5/2
by hooke's law i need to find the integral of 5/2x right. but what are the bounds.
All you have to do is use the formula
Work = potential energy increase = (1/2) k x^2, where k is the spring constant that you computed (which is 2.5 lb/inch) and x = 6 inches.
The answer wil be in inchlb. If you want it in foorlb, you will have to divide by 12.
You could also do it by integration
Work  Integral of F dx = Integral of kx dx from x=0 to x=6. The answer is the same.

15/4...... (1/2)(2.5)(6^2)=45 in/lb converto to feet divide by 12... 15/4
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