# Conservation of energy

1. May 12, 2014

### Karol

1. The problem statement, all variables and given/known data
A spring with a constant k=10[N/m] is attached with a long rope to a weight of 0.05 kg. they both lie horizontally on the table. the string is stretched 10 cm and released and the weight travels 40 cm on the table until it halts. what is the coefficient of friction

2. Relevant equations
Elastic energy of spring:$E_P=\frac{1}{2}kx^2$

3. The attempt at a solution
Conservation of energy:
$$\frac{1}{2}kx^2=mg\mu\cdot \triangle x$$
$$\frac{1}{2}10\cdot 0.1^2=0.05 \cdot 10 \cdot \mu \cdot 0.4 \rightarrow \mu=0.25$$

#### Attached Files:

• ###### Spring.jpg
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2. May 12, 2014

### Staff: Mentor

With friction present, the spring may end up still stretched a bit.

3. May 12, 2014

### Karol

I don't see how this can happen in this case since the spring gives velocity to the mass which continues travelling.
In any case the initial potential energy of the spring is the same, it is stretched 10 cm.

4. May 12, 2014

### Staff: Mentor

If the spring remains stretched a bit, it retains some potential energy.

5. May 12, 2014

### Karol

The spring will stop at a distance of:
$$F=mg\mu=kx\rightarrow 0.5\mu=10x \rightarrow x=0.05\mu$$
Conservation of energy:
$$\frac{1}{2}k \left(0.1^2-0.05^2\mu^2\right)=mg\mu\triangle x$$
$$5(0.01-0.0025\mu^2)=0.5\mu 0.4$$
$$0.0125\mu^2+0.2\mu-0.05=0 \rightarrow \mu=0.246$$
It still isn't μ=0.2

6. May 12, 2014

### rude man

Right now I don't see how the problem can be solved unless the coefficient of static friction is also given.

We start to pull on the spring. At first, the block stays put. At some spring stretch level the block starts to move. This will happen long before the full 10 cm of spring tension is reached since at 10cm the spring force is 1N, way more than what is required to start the weight moving even for a high static friction coefficient. So we have to continue pulling on the spring even when the weight has started to accelerate, until the full 10cm of spring tension is reached. So the puller applies not only the energy in a 10cm-expanded spring but also his force x distance until the 10 cm spring extension is reached.

That is the total system input power. The system output power is of course weight x kinetic friction coefficient x distance traveled after the spring is released.

A pretty complicated problem and I wonder if the wording is correct.

7. May 12, 2014

### Karol

The system isn't like that. you don't start to pull the spring until the mass also moves, no. you pull the mass which pulls the string through the rope. i knew my simple drawing isn't good enough.
You pull the mass, which stretches the spring, a distance of 10cm and release.

8. May 12, 2014

### Karol

Installation

There can be no residual elongation of the spring because of the installation. the spring is pulled by a long cord. at the end of the cord is the mass. the mass is released and it travels until it stops, in front of the spring.

#### Attached Files:

• ###### Snap1.jpg
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Last edited: May 12, 2014
9. May 12, 2014

### Staff: Mentor

Right. I hadn't looked at this closely enough. It' doesn't seem complicated. The spring gives up all its energy. The mass travels 40cm.

So your calculations seem right. μk = 0.255

Maybe one of the data values isn't right? e.g., say the mass travels 51cm ....

10. May 13, 2014

### rude man

OK, with this picture I don't see why your original answer isn't correct. Hope you can post what you found out later about it.

EDIT: I'll bet the spring traveled 40 cm from its original position. That makes Δx = 0.4 + 0.1 = 0.5 and then the answer is indeed μ = 0.2.

Last edited: May 13, 2014