Cart Striking Spring: Velocity & Equations

[Dustinsfl]

A cart starts from a height h and goes through a loop of radius r. After going through the loop the cart stikes a spring with spring constant k. How for is the spring depressed.

I have solved for the velocity at the top of loop, \(v^2 = gr\).
I am stuck with coming up with the equation of motion for the cart strike.
\[
m\ddot{x} = -kx +\text{veloctiy?}
\]
do I have \(\sqrt{gr}\) as well? If not, are there any other forces to consider?

 

[I like Serena]

A cart starts from a height h and goes through a loop of radius r. After going through the loop the cart stikes a spring with spring constant k. How for is the spring depressed.

I have solved for the velocity at the top of loop, \(v^2 = gr\).
I am stuck with coming up with the equation of motion for the cart strike.
\[
m\ddot{x} = -kx +\text{veloctiy?}
\]
do I have \(\sqrt{gr}\) as well? If not, are there any other forces to consider?

The equation of motion is:
\[
m\ddot{x} = -kx
\]
And no, there are no other forces to consider once the cart is coming out of the loop.

But rather than using the equation of motion, I would use that the energy of a compressed spring is $\frac 1 2 k x^2$, which must be equal to the initial gravitational energy $mgh$.

Btw, note that it is not given that the cart starts from the top of the loop. It is only given that it starts at height $h$.

 

[Dustinsfl]

The equation of motion is:
\[
m\ddot{x} = -kx
\]
And no, there are no other forces to consider.

But rather than using the equation of motion, I would use that the energy of a compressed spring is $\frac 1 2 k x^2$, which must be equal to the initial gravitational energy $mgh$.

Btw, note that it is not given that the cart starts from the top of the loop. It is only given that it starts at height $h$.

The height h is a point before the loop. \(h\geq \frac{5}{2}r\) in order to make it through the loop.

So you are saying use the conversation of energy equation.
\[
mgh_i+ 0\cdot KE = mgh_f + \frac{1}{2}kx^2
\]
What would \(mgh_f\) be though?

 

[I like Serena]

The height h is a point before the loop. \(h\geq \frac{5}{2}r\) in order to make it through the loop.

So you are saying use the conversation of energy equation.
\[
mgh_i+ 0\cdot KE = mgh_f + \frac{1}{2}kx^2
\]
What would \(mgh_f\) be though?

I assume $h$ is the height above the spring.

 

[Dustinsfl]

I assume $h$ is the height above the spring.

Yes. The spring would be h = 0 after the cart goes through the loop.

So is it
\[
mgh = \frac{1}{2}kx^2\Rightarrow x = \sqrt{\frac{2mgh}{k}}
\]
but using the fact that \(h\geq \frac{5}{2}r\), we have
\[
x\geq\sqrt{\frac{5mgr}{k}},
\]
correct?

 

[I like Serena]

Yes. The spring would be h = 0 after the cart goes through the loop.

So is it
\[
mgh = \frac{1}{2}kx^2\Rightarrow x = \sqrt{\frac{2mgh}{k}}
\]
but using the fact that \(h\geq \frac{5}{2}r\), we have
\[
x\geq\sqrt{\frac{5mgr}{k}},
\]
correct?

Yes.
That is assuming that the cart stays in contact with the loop while going through it.
It seems that that piece of information is not given.