# Compression of a spring with friction

1. Sep 5, 2012

### Scorpy

Hi again

I try to solve this by my own with little help from the net, but still I'm not sure if it's the correct!

Here is my work, and please someone tell me if I'm wrong, again. (x = 20,8 m ???)

Thanks

P.S. Here is the full problem http://i.minus.com/j8Fbhtd9CJtCQ.jpg [Broken]

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Last edited by a moderator: May 6, 2017
2. Sep 5, 2012

### voko

Some parts of the problem text are washed out, and the solution is out of focus. I don't think you want any guesswork on our part, so you should give us something more readable if you need help.

3. Sep 5, 2012

### Scorpy

A sledge of mass m=2 kg starts with initial velocity Va=7 m/s at ha=40m. Compute, neglecting friction.
a) the speed Vb at hb=20m

The sledge then travels from C (hc=30m) to D horizontally on a rough surface with uk=0.6.
From D the surface becomes frictionless. In E a spring in equilibrium condition (i.e. neither streched nor compressed) is positioned.
b) Determine the maximum compression of the spring, knowing that CD=20m and the spring constant k=0.2 N/cm.

4. Sep 5, 2012

### azizlwl

PE+KE=Wf+PEs

mgh + 1/2 mv2= mgμd + 1/2 kx2
Check for zero PE reference level.
Gravitational force is conservative, only levels are relevant not the path between the levels.

5. Sep 5, 2012

### Scorpy

So, I take Vc and d as CD?

6. Sep 5, 2012

### Scorpy

So, for v I take Vc and for d = CD?

7. Sep 5, 2012

Yes.

8. Sep 5, 2012

### Scorpy

And then, the final result is x=7.6m if I'm not wrong again?

9. Sep 5, 2012

### azizlwl

Check the unit of spring constant k.

10. Sep 5, 2012

### CAF123

I get a very similar answer(≈7.7m).

11. Sep 5, 2012

### azizlwl

mgh + 0.5mv2=mgμd+0.5ky2
2x9.8x10 +0.5x2x7x7=2x9.8x0.6x20 + 0.5x0.2/0.01 x y2
196+49 =235.2 +10xy2
9.8=10xy2

y2= 0.98
y=1m

12. Sep 5, 2012

### CAF123

Why is h = 10 m and v = 7m/s? I computed the velocity at C to be 15.7m/s and the motion is at a height of 30m. (I took the reference of zero potential at ground level).

13. Sep 5, 2012

### azizlwl

As i mentioned on previous reply, it is a conservative force. It is not he path but the different of height that determines the ΔPE. So the ΔPE is height between A and C.
Conservation of energy. The kinetic energy remain the same. As in projectile, the horizontal velocity remains constant.

From point A to C no energy dissipated.

add: If level A equal to level C, do you need to take zero reference at ground level?

14. Sep 5, 2012

### CAF123

Ah of course! But I still don't see why the velocity at C is 7m/s. Can you show this mathematically? I have tried but get values greater than this.

15. Sep 5, 2012

### PhanthomJay

There are too many steps here that cause errors. Eliminate all the intermediate steps. Start at A and end at E. $(KE + PE)_{initial- at- A} + W_f = (KE + PE)_{final- at -E}$
Work done by friction is numerically a negative value.

16. Sep 5, 2012

### Scorpy

So, guys, what is the final and also the right result?

17. Sep 5, 2012

### PhanthomJay

Oh heck you have a good understanding of work energy concepts, so it's just math now for the 2nd part and using g = 10 , then

1/2(2)(7)^2 + (2)(10)(40) + 0 - .6(2)(10)(20) = 0 + 2(10)(30) + 1/2(20)(x^2)

From which x = 1 m.

Which looks like what azizlwl got using a slightly different approach between a different set of points. You should confirm this on your own, as there are lots of ways to skin a cat, as they say.

Oh sorry I didn't pay attention to part A I guess that's solved already.

18. Sep 5, 2012

### Scorpy

Okay, the part a) is just fine, no problem at all with it...
Many thanks to everybody, I hope the answer of 1m is correct :P

19. Sep 5, 2012

### CAF123

Can anyone enlighten me as to why the velocity at C is 7m/s? Many thanks

20. Sep 5, 2012

### PhanthomJay

You should confirm it is correct on your own.....I didn't take the time pouring through your pages of calcs to see why you arrived at a different answer...maybe too many steps and a number got transposed somewhere... or maybe our answer is wrong you know.....