# Mass sliding on frictionless inclined plane compresses a spring

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1. Feb 26, 2017

### Nacho Verdugo

1. The problem statement, all variables and given/known data
A body with mass m start from repose and slide along a plane without friction. The plane forms an angle with the horizontal. After sliding a distance d unknown, the mass touches the extremity of a spring non compressed and non stretched with no mass. The mass slides a distance x until a momentary rest by the compression of the spring (of constant k). Find the initial separation d between the body and the extremity of the spring.

2. Relevant equations
So first, I tried to translate this exercise from Portuguese, so if some words doesn't make sense, I'm sorry!

The doubts I have is where to define the highness at zero. I tried to put z=0 at the moment when the mass compress the spring, in the momentary rest. Then I used conservation of energy and obtained the following result:

3. The attempt at a solution

Energy A (when the mass start to slides):

E=mg(d+x)sinβ

Energy B (when the mass compress the spring):

E=½kx^2

Considering that there's no friction
E_A=E_B

and I obtain that the distance d equals to:

d=½kx(x-1)

Is it right?

2. Feb 26, 2017

### kuruman

Hi Nacho Verdugo and welcome to PF

The idea is correct. Can you show how you got the answer? It should depend on the angle of the incline β.

3. Feb 26, 2017

### Nacho Verdugo

Hi! So, I just had second thoughts about what I post. The idea is to conserve the energy, right? So the energy in A is : mg(d+x)sinβ and the energy in B is :½kx^2. Conserving the energy I obtain:

mg(d+x)sinβ=½kx^2
⇒ mgdsinβ + mgxsinβ = ½kx^2
⇒ mdgsinβ = ½kx^2 - mgxsinβ
⇒ d= (kx^2)/(2mgsinβ)-x

I think in this solution the distance depends of the incline!

4. Feb 26, 2017

### PeroK

That looks good.

Last edited: Feb 26, 2017
5. Feb 26, 2017

### Staff: Mentor

Moderator note: Changed thread title to be descriptive of the problem: "Mass sliding on frictionless inclined plane compresses a spring"

6. Feb 27, 2017

### mastermechanic

But shouldn't it be $$d=\frac {kx^2} {2mgsinθ}$$? Why did you take into account x? The displacement x is traveled by both spring and box, so it is not necessary to add x to equation. Think about the moment that the box touches the spring, until that moment the box's energy will have been changed mgdsinθ and it will transfer this energy to the spring.Thus, it should be $$mgdsinθ =\frac 1 2(kx^2)$$ and we obtain $$d=\frac {kx^2} {2mgsinθ}$$

Is there anything I missed?

7. Feb 28, 2017

### PeroK

You missed that the box continues to lose gravitational PE as it compresses the spring, for the distance $x$.

8. Feb 28, 2017

### mastermechanic

But you missed as well that the lost gravitational PE is already being transfered in the moment.

Edit: I've just got it by thinking last position of them. There is no problem it should be $$d=\frac {kx^2} {2mgsinθ} - x$$

Last edited: Feb 28, 2017