# Physics Spring Problem

1. Mar 14, 2016

### Jamest39

1. The problem statement, all variables and given/known data
A 50-gram mass is accelerated from rest be a compressed spring (k = 1800 N/m), sending it on a journey along a frictionless loop-de-loop of radius 0.324 m. What minimum amount of initial compression of the spring is required if the mass is to remain in contact with the track at the top of the loop?

2. Relevant equations
centripetal acceleration = (v^2)/r
F = ma
F = kx (Hooke's Law)

3. The attempt at a solution
The velocity required at the top of the loop:
v = sqrt( 9.80m/s^2 * 0.324 m) = 1.782 m/s
Since the spring constant k is already known, we'd just need to determine F and use Hooke's Law to solve for x. But I can't figure out how to determine the force required to make push that mass and give it that velocity at the top of the loop.

2. Mar 14, 2016

### steven george

I would think about using the potential energy of the spring, 1/2 kx2. That amount should be equal to the kinetic energy for the speed you need added to the gravitational potential energy for a height equal to the diameter of the loop.

3. Mar 14, 2016

### Jamest39

So the gravitational potential energy is just mgh = (0.05 kg)(9.8 m/s^2)(0.648m)
But how can I solve for the potential energy of the spring without x (the amount of compression)?

4. Mar 14, 2016

### The Vinh

I have looked through your OP and I will show you how to find X. But before I tell you, have you learned popential energy and kinetic energy ? I don't want to write down an answer without your understanding.

5. Mar 14, 2016

### The Vinh

And does your homework provide the distance between the spring and the loop - de - loop ( sound like french)

6. Mar 14, 2016

### Jamest39

yes, we learned about it, I have the formulas for kinetic and potential energy. And it does not provide any distance, so I guess we can assume it starts right on the circle?

7. Mar 14, 2016

### The Vinh

Sound good, in order to let the substance remain contact with the track at the top of loop - de - loop, centripetal acceleration must be greater than gravity, Ac >= G. It means that V^2 = gr.
+) Frame of reference is the ground
+) 1/2V^2 + mg2r = 1/2Vo^2 => Vo^2 = gr + 2g2r
+) you use the conservation of energy: 1/2kx^2 = 1/2mVo^2 => you can find X ( i have to let you do the rest )

8. Mar 14, 2016

### The Vinh

For the spring, i choose the original length as frame of reference

9. Mar 14, 2016

### Jamest39

So then, using those, we would get:
x = sqrt( (10mgr)/k )

10. Mar 14, 2016

### The Vinh

That's right. And are you learning Pascal programming language ? because i saw sqrt()

11. Mar 14, 2016

### The Vinh

12. Mar 14, 2016

### Jamest39

Ah, I see. I dropped one of the 1/2 while working through it. Thanks!
And I'm not learning that programming language, that's just how I have to enter in square roots on my online math homework.

13. Mar 14, 2016

### The Vinh

I see

14. Mar 14, 2016

### steven george

If you know how much potential energy it has at the top, as you have described then just add the kinetic energy for the speed that you have already calculated. That will give you the total amount of energy. All of that energy comes from the potential energy of the spring so the total energy is equal to the elastic potential energy and "x" is the only thing that you don't know in the equation.