# Find Max Compression of a Spring

1. Jul 2, 2012

### Twiggy92

A 0.50 kg block is pushed against a 400 N/m spring, compressing it 22 cm. When the block is released, it moves along a frictionless horizontal surface and then up an incline (which has friction). The angle of the incline is 37 degrees and the coefficient of kinetic friction is 0.25. Use the conservation of energy law to find the maximum compression of the spring when the block returns to it.

Wother = ΔKE + ΔUg + ΔUel
KE = 0.5 * m * v2
Ug = m*g*y
Uel = -0.5 * k * compression2

This is actually a multi-step problem and it is the last part that I am stuck on. I have found the speed of the block just after it leaves the spring (6.2 m/s), the distance up the ramp that the block travels (2.47 m), and the height of the ramp where the block stops and begins to slide down again (1.48 m).

To find the maximum compression of the spring when it slides back down the ramp, I am trying to find the compression variable from the Uel equation, correct? I can solve for ΔUg, plug in k for the ΔUel equation and leave the compression as the variable I am trying to find. But for the ΔKE I don't know how to find the velocity. It is not the same as I found for when the block leaves the spring intially, right?

2. Jul 2, 2012

### PhanthomJay

Right. So what is the initial KE on its way down the slope, and what is its final KE when it compresses the spring and comes to a temporary stop?

3. Jul 2, 2012

### Twiggy92

If the velocity is 0, then the KE is 0. So the initial KE when the block is on its way down the slope is 0 because at the top of the slope the block has stopped to change direction. Is the final KE also 0 because it is temporarily stopped there?

4. Jul 2, 2012

### PhanthomJay

Yes, correct.

5. Jul 2, 2012

### azizlwl

On the way up
KEup=mgh+Ffx
On the way down
KEdown=mgh-Ffx

Last edited: Jul 2, 2012
6. Jul 3, 2012

### TSny

Twiggy92,

It's ok to work with the KE at the bottom of the slope. But you don't have to. To see the full power of energy concepts, you can skip the KE.

Let point A be the initial point where the mass is compressed against the spring a distance xo. Let point B be the point where the mass reaches its highest point on the incline, and let point C be the final point where the mass is compressed against the spring a distance xf.

Use Wother = ΔKE + ΔUg + ΔUel with point A as initial point and point B as final point. Note that KE is zero at both of these points. See if you can get the equation into the form

$\frac{1}{2}$kxo2 = mgdsinθ + μkmgdcosθ where d is distance traveled along slope.

Then apply Wother = ΔKE + ΔUg + ΔUel using B as the initial point and C as the final point. See if you can get

$\frac{1}{2}$kxf2 = mgdsinθ - μkmgdcosθ

If you then divide the second equation by the first, you can get an expression for xf2 / xo2. The result can be simplified by cancelling out common factors of m, g, and d.

However, if you feel more comfortable breaking up the problem into more parts by finding the KE at the bottom of the ramp going up and then going down, then go with that method.

7. Jul 4, 2012

### Zarrey

I have the same problem for my physics class, must be using the same book or something.

for this problem i got a different value for how far it goes up the slope. I got .885m, however i'm not sure this is right. I went through my logic multiple times and I still come up with my answer instead of yours.

As for the velocity I got the same result.

The compression when the object comes back down the slope is where i get lost. I'm not sure how to set up the conservation of energy theorem.

8. Jul 4, 2012

### TSny

Setting up the energy equation for coming back down is very similar to setting it up for going up the slope. See azizlwl's post. It would help if you could give us more detail on how you are setting up your equations.

I think Twiggy92's result of 2.47 m for the distance traveled up along the slope is correct.

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