A force of ten pounds compresses a 21 inch spring 8 inches. how much work is done in compressing the spring from a length of 15 inches to a length of 13 inches?

Question

anonymous · Answer

$$\int\limits_{13}^{15}(\frac{ 5 }{ 4 } x) dx$$ ???

anonymous · Answer

its not though the answer is 17.5 ft-lb... that gives 35 ft-lb, what am i doing wrong

anonymous · Answer

sorry 35 in-lb not ft-lb

perl · Answer

We know from hook's law 
F = k*d 

10 = k*8
10/8 = k 
5/4 = k

anonymous · Answer

yeah hookes law used it got that 5/4=k, I am stuck I have no clue how to get to 17.5 ft lb from that

perl · Answer

you have to consider the natural length of the spring

anonymous · Answer

so do I subtract an integral for 15 to 21?

perl · Answer

$$
\Large \int_{21-15}^{21-13}\frac{ 5 }{ 4 } x~ dx

$$

anonymous · Answer

lol awesome I was pulling my hair out, that will not be forgotten... So is that 17.5 now in ft lb or still in inch lb?

perl · Answer

we are in inch*lbs because of how we found k (using inches and pounds)

anonymous · Answer

thanks you perl

perl · Answer

$$

\Large \int_{21-15}^{21-13}\frac{ 5 }{ 4 } x~ dx = 
\Large \int_{6}^{8}\frac{ 5 }{ 4 } x~ dx = 17.5 ~~inch \cdot lbs

$$