Mastering Physics Solutions: Exercise 5.25

Mastering Physics Solutions: Exercise 5.25

On December 24, 2011, in Chapter 05: Work and Energy, by Mastering Physics Solutions

Exercise 5.25

Part A = 1.9J Click to use the calculator/solver for this part of the problem
Part B = 2.9J Click to use the calculator/solver for this part of the problem

Solutions Below:

A particular spring has a force constant of 1.5×103 N/m.

Part A

How much work is done in stretching the relaxed spring by 5.0 cm?
Express your answer using two significant figures.

Spring potential energy is given by U = 1/2kx2. When compressing or stretching an ideal spring, all of the work done to compress/stretch it is stored as potential energy. So:

U = 1/2kx2
U = 1/2(1500)(0.05)2

U = 1.9J

Part B

How much more work is done in stretching the spring an additional 3.0 cm?

Note that finding the potential energy for a stretch of 0.03m won’t give us the right answer here – since x is squared, we need to take into consideration all of the stretch/compression that was already done. So we can solve for the potential energy after 0.08m and then subtract the potential energy from the first 0.05m:

U = 1/2kx2
U = 1/2(1500)(0.08)2
U = 4.8J

Now subtract the 1.9J from Part A:

4.8 – 1.9 = 2.9

So the answer is 2.9J

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