integrals and springs

[mathstresser]

A force of 10 lb is required to hold a spring stretched 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?

I use the equation f=kx. Thus 10=4k. So, k=5/2.

Then, to find the work, I do the integral from 0 to 6 of (5/2)x.

So I get (5/4)x^2 evaluated from 0 to 6. The answer I come up with is 45. But the answer should be 15/4. What am I doing wrong?

 

[galactus]

Are you sure that 15/4 is correct. It takes 10 lbs to stretch it 4" beyond its natural length, One would think it's going to take more than that to stretch it 6" beyond its natural length. 15/4 is only 3.75.

 

[Daniel_Feldman]

I'm pretty sure the answer is 45.

If it takes 10 lbs to hold it stretched 4 inches, then by Hooke's Law F=kx >>>k=F/x=10/4=5/2

So the work required to stretch it 4 inches is (1/2)kx^2=(1/2)(5/2)(4)^2=20 inch-lbs.

The work to stretch it an additional 2 inches is the integral, from 4 to 6, of (5/2)x dx.

This equals (1/2)(5/2)x^2 from 4 to 6 =(5/4)x^2 from 4 to 6=(5/4)(36)-(5/4)(16)=45-20=25.

So the total work done is 20+25=45 inch-lbs.

 

[galactus]

I agree Daniel. Another case of an erroneous textbook answer?.

 

[mathstresser]

They're both right.

It's 45 in-lb (if that exists)

and 15/4 ft-lb

Thank you for your help, but I'm sorry you wasted your time.