392,799 views
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
by
2.2k points

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
by
2.8k points