# Work Done by a spring

1. Feb 20, 2004

### AngelofMusic

Hello!

Sorry I've got so many questions lately. Studying for a midterm right now. Anyway, I've been trying to solve problems involving springs, and I'm getting confused about calculating the work done by a spring when its initial state is already in compression.

The question: A 3-kg block rests on top of a 2-kg block supported by but not attached to a spring with k = 40N/m. The upper block is suddenly removed. Determine the maximum velocity and the maximum height of the 2-kg block.

I figured that the maximum velocity would occur before the block flew off the spring, and probably before it returned to its unstretched length.

In equation: T1 + U1->2 = T2

$$U1->2 = 1/2k(x1)^2 - 1/2k(x2)^2$$, where x1 and x2 are measured from the unstretched length of the spring.

I calculated x1 = 1.22625 m. So:

$$0 + 30.07 J - 1/2k(x2)^2 = 1/2mv^2$$

Taking the derivative and setting it to zero:

-k(x2)=0, which isn't correct. How am I supposed to approach this problem? I think my equation for work done by the spring is incorrect, but I don't know what it is supposed to be.

2. Feb 20, 2004

### NateTG

There's a couple of options:

A clever option is to realize that the velocity is maximal when the net force on the block is zero, so when $$mg=kx$$ and then use energy.

Alternatively, you can calculate the Kinetic Energy as a function of position:
$$E_k=\frac{1}{2}kx_0^2-\frac{1}{2}kx^2-mg(x-x_0)$$
then take the derivative and set it to zero since the kinetic energy is maximal when the velocity is maximal.
$$kx-mg=0$$
so
$$mg=kx$$ (have I seen this equation before?)
$$x=\frac{mg}{k}$$

You could also calculate an expression for the velocity, take a derviative, and then work your way back, but it's not nearly as nice in this case:
$$v=\sqrt{\frac{2}{m}(\frac{1}{2}kx_0^2-\frac{1}{2}kx^2-mg(x-x_0))$$
for velocity as a function of position.

3. Feb 20, 2004

### AngelofMusic

Just a quick follow-up on the signs of the equations. When you have:

$$E_k=\frac{1}{2}kx_0^2-\frac{1}{2}kx^2-mg(x-x_0)$$

Wouldn't the result be:

-kx - mg = 0

This is always a confusing point for me. I never know when the spring force is considered negative/positive, and whether the work done is negative or positive.

In this case, the spring is going back towards its original position, so I'd say it's positive. How does that translate numerically into the equation?

$$E_spring = \frac{1}{2}kx_0^2 - \frac{1}{2}kx^2$$

Since it was originally compressed by 1.22625 m, $$x_0$$ = -1.22625 and x = 0.4905 when I set mg = kx. Since the values are squared, though, I'm still getting a negative since the x values are sqared.

4. Feb 20, 2004

### NateTG

Oh the maximum velocity is definitely at:
$$x=-\frac{mg}{k}$$
since the other solution with $$x>0$$ is not in the domain where the function is meaningfull.

I had a sign error. Good question/catch.

I'm not sure if you've seen work integrals but essentially, the work can be described as
$$W=\int \vec{F}_{net} \dot{} v dt$$
and if you use the same coordinate system for everything, you'll get the correct sign.

Alternatively, work done by forces in the same direction as the velocity is positive, and work done by forces opposing the velocity is negative.

Regading the math, even with the sign error:
$$(x_0^2-x^2) > 0$$

Here's what I get using approximations:
$$x_0=-1.2$$ (Negative since down is negative)
$$x_{vmax}=-.5$$
now
$$x_0^2=1.44$$
$$x_{vmax}^2=.25$$
so
$$v_{max}=\frac{2}{2} \sqrt{\frac{1}{2}40(1.19) - mg(-.7)}$$
$$v_{max}=\sqrt{23.8-14}=\sqrt{10}$$
so I get
$$v_{max}=3$$
or so.