Exercise 5.25

Solutions Below:

A particular spring has a force constant of 1.5×10

^{3}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.

Express your answer using two significant figures.

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

U = 1/2kx^{2}

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 = 1/2kx^{2}

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