A spring has natural length 25 cm. Compare the work W1 done in stretching the spring from 25 cm to 35 cm with the work W2 done in stretching it from 35 to 45 cm.

Question

A spring has natural length 25 cm. Compare the work W₁ done in stretching the spring from 25 cm to 35 cm with the work W₂ done in stretching it from 35 to 45 cm. (Use k for the spring constant)

w1=?

w2=?

How are W₂ and W₁ related?

Answer 1

Unless I am reading the question wrong, I agree with Mark's conclusion.

As anyone who has gone to a gym has experienced (or has worked with springs and elastomers), pulling on a spring is initially easy but gets harder with each incremental increase of stretch.

Answer 2

When a constant force F is applied throughout a distance d, the amount of work done is

W = Fd

This makes sense, because if you double the force (you pull twice as hard), you're doubling the amount of work done.

And if you double the distance (you drag the box 2 miles instead of 1 mile), you're doubling the amount of work done.

In the first part of this problem, the distance is 10 centimeters (because 35 cm - 25 cm = 10 cm). But the force that we're applying throughout this distance is not constant: we have to pull harder and harder as the spring gets stretched.

If you apply a force to lengthen the spring a very small amount, say from length x to length x + Δx, the force does have to increase as the spring gets longer. But the force increases only a very small amount over this very small length: the force stays almost constant and stays almost exactly equal to kx (where k is the spring constant) throughout this very small length. We can therefore approximate the work we did by using our formula

W = Fd

In this formula, the distance d is Δx, and the force F is kx.

So the amount of work it takes to stretch the spring from length x to length x + Δx is approximately

Fd = (kx)(Δx).

Now we can use Calculus and change the above kx Δx to kx dx.

The amount of work it takes to stretch the spring from its natural length to 10 cm longer is exactly

∫ (from x=0 to x=10) kx dx = k(1/2)x² (from x=0 to x=10) = (k/2)10² - (k/2)0² = 50k = W₁.

And the amount of work it takes to stretch the spring from its natural length to 20 cm longer is exactly

∫ (from x=0 to x=20) kx dx = k(1/2)x² (from x=0 to x=20) = (k/2)20² - (k/2)0² = 200k.

Therefore the amount of work it takes to stretch the spring from 10 cm longer than its natural length to

20 cm longer than its natural length is 200k - 50k = 150k = W₂.

So the answer is: W₂ is 3 times as great as W₁ because 150k is 3 times as great as 50k. They never told us what units of force were employed, or what the value of k is, but it's still true that W₂ is 3 times as great as W₁.

Answer 3

we can use the formula for the work done in stretching a spring:

W= 1/2 * k * (Δx)²

where:

1. W is the work done,
2. k is the spring constant,
3. Δx is the change in length of the spring.

Given that the natural length of the spring is 25 cm, we can calculate the changes in length:

For W₁​, stretching from 25 cm to 35 cm: Δx₁​=35cm−25cm=10cm

For W₂​, stretching from 35 cm to 45 cm: Δx₂​=45cm−35cm=10cm

Now, let's calculate the work done for each case:

W₁​=1/2 * k * (10cm)² = k * 50cm²

W₂​=1/2 * k * (10cm)² = k * 50cm²

We can see that both W₁ and W₂ are equal, regardless of the specific value of k as the spring is stretched over an equal displacement in both cases.