Mass sliding on frictionless inclined plane compresses a spring

[Nacho Verdugo]

Homework Statement

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.

Homework 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:

The Attempt at a Solution

[/B]
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?

 

[kuruman]

Hi Nacho Verdugo and welcome to PF

and I obtain that the distance d equals to:

d=½kx(x-1)

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

 

[Nacho Verdugo]

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.

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!

 

[PeroK]

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!

That looks good.

 

[gneill]

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

 

[mastermechanic]

That looks good.

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?

 

[PeroK]

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?

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

 

[mastermechanic]

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

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

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