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34 votes
34 votes
A spring with spring constant 39N/m hangs at rest. A force then pulls the spring causing it to stretch from 0.3m to 0.9m. How much work was done by the force pulling on the spring

User Andre Yonadam
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1 Answer

11 votes
11 votes

Answer:

14.04 J

Step-by-step explanation:

First, we need to calculate the magnitude of the force, so by Hooke's law, it is equal to

F = kx

Where k is the spring constant and x is the stretch. Since it goes form 0.3 m to 0.9 m, the value of x is:

x = 0.9 m - 0.3 m

x = 0.6 m

Then, replacing k = 39 N/m and x = 0.6 m, we get that the force is

F = (39 N/m)(0.6 m)

F = 23.4 N

Now, the work done is equal to the force times the distance, so

W = Fx

W = (23.4 N)(0.6 m)

W = 14.04 J

Therefore, the work done by the force was 14.04 J

User Thinkbigthinksmall
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