# Compression of a spring with friction

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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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.

Scorpy
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.

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.

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.

Scorpy
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.

So, I take Vc and d as CD?

Scorpy
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.

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

Gold Member
So, for v I take Vc and for d = CD?
Yes.

Scorpy
Yes.

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

azizlwl
Check the unit of spring constant k.

Gold Member
And then, the final result is x=7.6m if I'm not wrong again?
I get a very similar answer(≈7.7m).

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

Gold Member
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
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).

azizlwl
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).

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?

Gold Member
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?
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.

Homework Helper
Gold Member
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.

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

Homework Helper
Gold Member
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.

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

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

Homework Helper
Gold Member
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... azizlwl
Can anyone enlighten me as to why the velocity at C is 7m/s? Many thanks

At point A only energy for the system is kinetic energy with velocity of 7m/s.
From point A to point C, gravity has done work on the sledge equal to mgh.
No energy is taken out from the sledge ONLY added work by gravity.

So at Point C we have original energy PLUS energy added by gravity.

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Gold Member
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... I think you think I am the OP :) I just came in and did the question and azizlwl pointed out my error in the way in which I was forgetting to use potential energy. However, I am still unsure of why the velocity at C is 7m/s. Computing it gives a much larger value of about 25m/s. I computed it like: $$\frac{1}{2}mv_b^2 + mg(h_A - h_B) = \frac{1}{2}mv_c^2 + mg(h_C-h_B),$$ which when using $v_b$= 21m/s yields $v_c$ as ≈25m/s. (reference of potential, 0 at ground)
See any errors?
Thanks

Homework Helper
Gold Member
Can anyone enlighten me as to why the velocity at C is 7m/s? Many thanks
The velocity at C is what you got...15 something m/s...velocity at A is 7 m/s...

Gold Member
The velocity at C is what you got...15 something m/s...velocity at A is 7 m/s...
$v_c =$ 15.7m/s was what I got when I did my potential energy contributions incorrect. I now get $v_c =$ 25m/s if my equation is correct that I posted in my previous post.

Homework Helper
Gold Member
I think you think I am the OP :) I just came in and did the question and azizlwl pointed out my error in the way in which I was forgetting to use potential energy. However, I am still unsure of why the velocity at C is 7m/s. Computing it gives a much larger value of about 25m/s. I computed it like: $$\frac{1}{2}mv_b^2 + mg(h_A - h_B) = \frac{1}{2}mv_c^2 + mg(h_C-h_B),$$ which when using $v_b$= 21m/s yields $v_c$ as ≈25m/s. (reference of potential, 0 at ground)
See any errors?
Thanks
Yes, if you are referencing the ground as 0 PE, then its mg (20) for initial PE at B, and mg(30) for final PE at C. I think you had it right the first go around, Vc is about 15 + m/s.....

azizlwl
I think you think I am the OP :) I just came in and did the question and azizlwl pointed out my error in the way in which I was forgetting to use potential energy. However, I am still unsure of why the velocity at C is 7m/s. Computing it gives a much larger value of about 25m/s. I computed it like: $$\frac{1}{2}mv_b^2 + mg(h_A - h_B) = \frac{1}{2}mv_c^2 + mg(h_C-h_B),$$ which when using $v_b$= 21m/s yields $v_c$ as ≈25m/s. (reference of potential, 0 at ground)
See any errors?
Thanks

Energy at A =KEA
Energy at B=KEA + ΔPEAB
Energy at C=KEA + ΔPEAB -ΔPEBC
Energy at C=KEA + ΔPEAC

Error in your total energy at Point B. v=vA not vB and vC should be vB

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Gold Member
Yes, if you are referencing the ground as 0 PE, then its mg (20) for initial PE at B, and mg(30) for final PE at C. I think you had it right the first go around, Vc is about 15 + m/s.....
Many thanks. I think the reason I might have got confused is because azizlwl might have used a different reference of potential.
Ok, so using the velocity at C to be 15.7m/s, i have: $$\frac{1}{2}mv_c^2 + mgh_c = \frac{1}{2}kx^2 + mg\mu(CD)$$. Putting in numbers gives x ≈ 7.7m?
Where is my error?

Gold Member
Energy at A =KEA
Energy at B=KEA + ΔPEAB
Energy at C=KEA + ΔPEAB -ΔPEBC
Energy at C=KEA + ΔPEAC

Error in your total energy at Point B. v=vA not vB
Surely I have some potential energy at A if my reference of potential is the ground?

azizlwl
Surely I have some potential energy at A if my reference of potential is the ground?

Potential energy at a point is meaningless.
It is only relative to other point that count.
When you move to other point, then only either energy is added or reduced.

Energy =FxD
If you apply force, here it not moving, means no energy added or taken out even gravity is available all the time.

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Gold Member
Potential energy at a point is meaningless.
It is only relative to other point have a value.
I agree. I measure the potential at various points of the journey with respect to the ground/ relative to the ground.

azizlwl
I agree. I measure the potential at various points of the journey with respect to the ground/ relative to the ground.

Maybe we approach different way.
If force is applied for a distance then the body gain energy equal to Fx
Now if gravity do work and PULL the object down, energy gain by the object.
The object gain energy depend on the distance it moved NOT from where(assuming g constant)

Does a point A higher than B have different ΔPE if moved equal distance?
Is position determine the energy equation?

The motion is between Point A, B and C only.
Making ground as reference will create unnecessarily messy calculation.

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Ok, so using the velocity at C to be 15.7m/s, i have: $$\frac{1}{2}mv_c^2 + mgh_c = \frac{1}{2}kx^2 + mg\mu(CD)$$. Putting in numbers gives x ≈ 7.7m?