# Stretching a stretched spring/non-stretched spring

1. Feb 14, 2010

### lemon

Hi:
1. Why is more work needed to stretch a spring by 1cm from its unextended length, than is required to stretch it by an extra 1cm when the spring is already extended?

Up until the elastic limit the spring obeys Hooke's law and the force applied is equivalent to the extension. After the elastic limit a greater force is required to stretch the spring for the same length as would be possible before the elastic limit.
What actually goes on inside the spring to make this true is anybody's guess.

Hazard a try?

2. Feb 14, 2010

### N-Gin

Take a look at the potential energies.

When we stretch a spring by a certan amount $$x$$ from unextended state. We need amount of work

$$W_{1}=E_{p_{1}}-0=\frac{1}{2}kx^2$$

where $$k$$ is spring constant.

Now, let's say that we stretch the spring again by $$x$$. But, we already have some potential energy in the system and the new length of the spring is $$2x$$.

$$W_{2}=E_{p_{2}}-E_{p_{1}}=\frac{1}{2}k(2x)^2-\frac{1}{2}kx^2=\frac{3}{2}kx^2$$

Obviously $$W_{1}<W_{2}$$.

So, we need more work to extend an already extended spring for the same amount. Of course, this is because the force linearly grows with the length of the spring.

Cause of such behaviour of springs is in the straining of the bonds between atoms in spring (atomic layers to be exact).

3. Feb 15, 2010

### lemon

N-Gin - simply thanks :)