What is the answer?

Question

A spring with a natural length of 3.0 meters is held at a length of 3.5 meters by a force of 5 newtons. How much work is done in stretching this spring from a length of 4.1 meters to a length of 4.6 meters?

Answer 1

Hi Kyler!

The force applied by a spring is k*x where k is the spring constant and x is the distance from it's rest position the spring is stretched. We know the rest position and the force applied, so we can solve for k:

5N = k*0.5m

k = 10

We could use some calculus to use force to obtain work using FxD, or we could do it an easier way. The potential energy contained within a stretched or compressed spring is 1/2*k*x^2. So if we take the difference in the energy stored in the spring at the start and end location (relative to the resting position) we will also obtain the work required to move the spring between those two locations:

1/2*10*1.5^2 - 1/2*10*1.1^2 = 5.2J

Edit:

This should have been

1/2*10*1.6^2 - 1/2*10*1.1^2 = 6.75J

Hope that helps!

Answer 2

A spring with a natural length of 3.0 meters is held at a length of 3.5 meters by a force of 5 newtons. How much work is done in stretching this spring from a length of 4.1 meters to a length of 4.6 meters?

SOLUTION

F= ks is Hooke's Law, which says the force required to displace a spring from equilibrium is proportional to the distance it is stretched.

"A spring with a natural length of 3.0 meters is held at a length of 3.5 meters by a force of 5 newtons."

This allows us to solve for the constant k that appears in Hooke's Law. F = 5 N, and s = 3 meters, so

5 = k * 3 => k=5/3 Newtons/meter.

Thus, F = 5/3 s, Work is Force times distance. Work is force times distance, so work is given by

W = Fs = (5/3 s) s = 5/3s².

We can now answer the question: How much work is done in stretching this spring from a length of 4.1 meters to a length of 4.6 meters?

CORRECTION

Thanks, Wyatt! I wasn't thinking about the fact that the work is a variable function of s!

Now we can integrate 5/3 s² ds from 0 to 1/2 = 5/9 s³ | from 0 to 1/2 = 5/9 (1/8) = 5/72 Newton-meters.