(b) How much work is done in compressing the springs? This one has been kicking

Question

(b) How much work is done in compressing the springs?

This one has been kicking my butt for several days as well as the butts of all the responders on Chegg so far. Although it's not really advanced considering this is question is straight out of a college freshman level physics book I've posted this one here in the Advanced Physics subject in hopes of better luck. So for five stars the responder must do at least one or more of the following:

1. Answer the question correctly.

2. Provide a detailed step by step solution with alternate numbers.

3. Provide a detailed step by step cookbook recipe for solving using the varibles accepted letters/symbols.

Thanks!

Explanation / Answer

I don't know how much of a cookbook I can give you but I can walk through the problem.

Think of the force as applied gradually, it won't matter to the ultimate result and makes the reasoning clearer. When the force has increased to 2.45 * 105 N, the leaf spring has compressed 0.5 m and is about to touch the helper spring.

The work done at this point is equal to the potential energy of the system, ½kx2

= (0.5)(4.9 * 105 N/m)(0.5 m)2 = 6.13 * 104 J

Now the spring constant has suddenly jumped up to the sum of the two spring constants, 8.7 * 105 N/m. We are going to apply an additional 2.05 * 105 N of force (to bring the total force up to 4.5 * 105 N), which will compress this combined spring by 0.236 m, for a total compression of 0.736 m, which is the answer to part (a).

The work done is trickier, we can no longer just depend on the potential energy calculation because of the nonlinear spring. So we have to integrate: W = F dx = kx dx = ½k(xfinal2 - xinit2)

= (0.5)(8.7 * 105 N/m)(0.7362m - 0.52m) = 1.27 * 105 J

for total work done of 1.88 * 105 J.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.