# Spring question

1. Apr 28, 2006

### plutonium

an object is connected to the wall through a spring and is resting at its equilibrium position. its mass is M and the coefficient of kinetic friction is 1/4. the object is flicked so that it will return to its original position (not instantly, but its oscillation is damped slowly).

what is the object's initial velocity? (you have to express it in terms of the other variables)

2. Apr 28, 2006

### Andrew Mason

I assume the object is resting on a horizontal surface and the spring is connected to the wall horizontally. The object is then given a brief impulse that causes it to move away from the wall.

If that is your question, the initial velocity does not depend on the co-efficient of kinetic friction or the spring. The answer to your question depends on the impulse.

$$v = F\Delta t/M$$

AM

3. Apr 28, 2006

### plutonium

ok here's the exact question:

a particle of mass m rests upon a rought horizontal plane with a coefficient kinetic friction of mu = 1/4 and is connected by a light elastic string with spring constant k. if the string is just taut find the initial speed, v, that the particle needs if it is projected directly away from P such that it just returns to its initial position.

$$a) g\sqrt{\frac{m}{2k}} b) g\sqrt{\frac{6m}{k}} c) g\sqrt{\frac{2m}{k}} d) g\sqrt{\frac{m}{k}} e) g\sqrt{\frac{k}{m}}$$

btw, is latex unique to this forum, or is there a program that i can download to make it work with ms word?

Last edited: Apr 28, 2006
4. Apr 28, 2006

### Andrew Mason

Now it makes sense.

How much energy does it have with initial speed v0?

How much energy does it lose in travelling a distance x?

How is the potential energy at maximum extension related to the initial Kinetic energy?

I don't think any of those answers is right

AM

5. Apr 29, 2006

### plutonium

$$\frac{1}{2} mv^2$$

well, the kinetic energy is converted into elastic potential and work done by friction.
$$\frac{1}{2} mv^2 = \frac{1}{2} kx^2 + \frac{mg}{4} x$$

then as it reaches its max stretch, it returns to its equilibrium point it turns into this:
$$\frac{1}{2} kx^2 = \frac{1}{2} mv^2 + \frac{mg}{4} x$$
(x1 and x2 should be the same)

but it will then go the other way and keep repeating itself until it loses all its initial kinetic energy to friction, so i thought it would be
$$\frac{1}{2} mv^2 = \frac{mg}{4} x$$
so wouldn't x in this case be the total distance (not displacement as it would be 0) it travels during the motion? how would i figure that out?

Last edited: Apr 29, 2006
6. Apr 29, 2006

### Hootenanny

Staff Emeritus
MASSIVE HINT: Energy is an undamped SHM system is conserved. I agree with AM, in my opinion non of the given answers are correct.

~H

7. Apr 29, 2006

### Andrew Mason

Correct. Since we know that it loses the rest of its energy in moving back to the equilibrium position, you can figure out the potential energy of the spring at maximum extension and from that work out the maximum extension.

Set out the equations that describe the energy of the system at maximum extension and again at 0. Then work out the solution for v.

AM

Last edited: Apr 29, 2006
8. Apr 30, 2006

### plutonium

Don't I have to work out the entire motion? Or can I just calculate it by assuming it moves back to its equilibrium position after just one extention?

9. Apr 30, 2006

### Andrew Mason

Isn't that the whole motion?

AM

10. May 1, 2006

### plutonium

By whole motion, I meant it vibrates for a while (hence covering more distance) before coming to a stop at its starting point. Wouldn't the negative work done by friction change as the mass moves vibrate more?

11. May 2, 2006

### Andrew Mason

The question says it does half a cycle only. Out and back. No vibration. What does the work done by friction depend on?

AM