# PhysicsHolt Physics, 2012 Ed., p. 183, #49: Concerning a Compressed Horizontal Spring with Friction

#### Ackbach

##### Indicium Physicus
Staff member
Problem: A light horizontal spring has a spring constant of $105 \; \text{N/m}$. A $2.00\; \text{kg}$ block is pressed against one end of the spring, compressing the spring $0.100 \; \text{m}$. After the block is released, the block moves $0.250 \; \text{m}$ to the right before coming to rest. What is the coefficient of kinetic friction between the horizontal surface and the block?

My attempt: Since there is friction, Conservation of Mechanical Energy does not apply. The Work-Energy Theorem does, however: $W_{ \text{net}}= \Delta KE$. Let us label the compressed state 1, the natural length of the spring state as 2, and the stopped state as 3. Indices will correspond to these states. If we examine the transition from 1 to 2, then
$$W_{ \text{net}}=KE_{f}= \frac{1}{2} mv_{2}^{2}.$$
But $W_{ \text{net}}=W_{el}-W_{f_{12}}$, so
$$W_{el}= \mu_{k}mgx_{12}+ \frac{1}{2} mv_{2}^{2}.$$
Here $W_{f_{12}}$ represents the work friction does in moving from 1 to 2, and $x_{12}$ is the distance from 1 to 2.

If we examine the transition from 2 to 3, we have that
$$-W_{f_{23}}=- \frac{1}{2} mv_{2}^{2} \quad \implies \quad \mu_{k} mgx_{23}= \frac{1}{2} mv_{2}^{2}.$$
Therefore,
$$W_{el}= \mu_{k}mgx_{12}+ \mu_{k}mgx_{23}= \mu_{k}mgx_{13}.$$
Hence,
$$\mu_{k}= \frac{W_{el}}{mgx_{13}}.$$
Now $W_{el}$ is an unknown. The question is, how can I compute $W_{el}$ without resorting to
$$W_{el}=\int \vec{F}_{el} \cdot d \vec{r} \; ?$$
If you compute this integral, you'll see that
$$W_{el}= \frac{1}{2} kx_{12}^{2}.$$
And if I carry out the rest of the algebra, I get
$$\mu_{k}= \frac{k x_{12}^{2}}{2mgx_{13}}= \frac{ (105 \; \text{N/m}) (0.1 \; \text{m})^{2}}{2 (2.0 \; \text{kg}) (9.8 \; \text{m/s}^{2}) (0.25 \; \text{m})} = 0.107,$$
which agrees with the book's answer.

I was thinking you might be able to do $W_{el}=- \Delta PE$, but doesn't that depend on the Conservation of Mechanical Energy? The other issue with using $W_{el}=- \Delta PE$ is that Holt Physics does not contain this equation.

So, just to be clear: my question is, "How can you solve this problem without using calculus?" It appears to me that friction is operational when the spring is pushing the block, so you can't just use conservation of mechanical energy for 1 to 2, and continue on from there.

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#### Opalg

##### MHB Oldtimer
Staff member
I don't know Holt's book, but I would be surprised if it doesn't somewhere give the formula $W=\frac12kx^2$ for the work done to compress a spring with constant $k$ through a distance $x$ (or equivalently the work done by the spring when it is released). The formula is included in the Wikipedia page on Work, together with a derivation which inevitably uses a bit of calculus – I don't see how that could be avoided.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
There's no need to involve the velocity in state 2.

The initial energy is:
$$\displaystyle \qquad E_{el} = \frac 1 2 k x_{12}^2 \qquad (1)$$
According to Conservation of Energy, this must be equal to the expended energy which is the work done by friction:
$$\displaystyle \qquad W_{friction} = F_{friction} x_{13} = \mu_k mg x_{13} \qquad (2)$$

Therefore:
$$\displaystyle \qquad E_{el} = W_{friction}$$
$$\displaystyle \qquad \frac 1 2 k x_{12}^2 = \mu_k mg x_{13}$$
$$\displaystyle \qquad \mu_k = \frac{\frac 1 2 k x_{12}^2}{mg x_{13}} \qquad (3)$$

I guess it depends on whether you can use (1) to decide if you need calculus (as Opalg remarked).
But applying Conservation of Energy really depends on having the formulas for the various energies and in particular elastic energy.

#### Ackbach

##### Indicium Physicus
Staff member
I don't know Holt's book, but I would be surprised if it doesn't somewhere give the formula $W=\frac12kx^2$ for the work done to compress a spring with constant $k$ through a distance $x$ (or equivalently the work done by the spring when it is released). The formula is included in the Wikipedia page on Work, together with a derivation which inevitably uses a bit of calculus – I don't see how that could be avoided.
Well, with a force that varies linearly, you can use the average value and use that to derive that $\langle F \rangle= k x/2$, and then multiply by $x$ to get the work done. But you're still relying implicitly on calculus.

What we are able to use is that the elastic potential energy is $k x^{2}/2$.

There's no need to involve the velocity in state 2.

The initial energy is:
$$\displaystyle \qquad E_{el} = \frac 1 2 k x_{12}^2 \qquad (1)$$
According to Conservation of Energy, this must be equal to the expended energy which is the work done by friction:
You're using a more robust version of the Conservation of Energy, right? Perhaps
$$W_{NC}= \Delta KE+ \Delta PE,$$
where $W_{NC}$ is the work done by any non-conservative forces? I've been thinking along these lines - the difficulty is that this more general work-energy theorem isn't mentioned in Holt Physics. I'm just trying to get my head around how my students, with primarily the Holt Physics textbook as their resource, are going to be able to solve the problem. And since this is an algebra-based physics course, the (admittedly most appropriate) tools of calculus are not available.

The most sensible thing would be to look up the solution in the solutions manual - problem is I don't know if I have access to that right now.

$$\displaystyle \qquad W_{friction} = F_{friction} x_{13} = \mu_k mg x_{13} \qquad (2)$$

Therefore:
$$\displaystyle \qquad E_{el} = W_{friction}$$
$$\displaystyle \qquad \frac 1 2 k x_{12}^2 = \mu_k mg x_{13}$$
$$\displaystyle \qquad \mu_k = \frac{\frac 1 2 k x_{12}^2}{mg x_{13}} \qquad (3)$$

I guess it depends on whether you can use (1) to decide if you need calculus (as Opalg remarked).
But applying Conservation of Energy really depends on having the formulas for the various energies and in particular elastic energy.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
the difficulty is that this more general work-energy theorem isn't mentioned in Holt Physics.
That's a pity.
It sounds rather limiting.
Especially since conservation of energy is one of the fundamental laws of nature that is applicable without any conditions.
Somehow I feel that should be mentioned to students.

#### Ackbach

##### Indicium Physicus
Staff member
I finally looked up the book's solution. It runs like this:

$$W_{ \text{net}}= \Delta KE=KE_{f} - KE_{i}= -KE_{i}= -PE_{g}= - \frac{1}{2} kx^{2}.$$
(Note: yes, it does say $-PE_{g}$, which stands for gravitational potential energy. I think they meant $PE_{e}$ for elastic potential energy.)
$$W_{ \text{net}}= F_{k}d \cos( \theta)= \mu_{k} mgd \cos( \theta)- \mu_{k}mgd.$$
$$\frac{1}{2} kx^{2}= \mu_{k}mgd$$
$$\mu_{k}= \frac{kx^{2}}{2mgd} = \frac{(105 \; \text{N/m})(-0.100 \; \text{m}^{2})}{2(2.00 \; \text{kg})(9.81 \; \text{m/s}^{2}) (0.250 \; \text{m})}= \boxed{0.107}.$$

So if I analyze this solution, I guess the step that pops out at me immediately is this: how can they know that $-KE_{i} = -PE_{e}$ (assuming it was a typo above, as I already mentioned)?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I finally looked up the book's solution. It runs like this:

$$W_{ \text{net}}= \Delta KE=KE_{f} - KE_{i}= -KE_{i}= -PE_{g}= - \frac{1}{2} kx^{2}.$$
(Note: yes, it does say $-PE_{g}$, which stands for gravitational potential energy. I think they meant $PE_{e}$ for elastic potential energy.)
$$W_{ \text{net}}= F_{k}d \cos( \theta)= \mu_{k} mgd \cos( \theta)- \mu_{k}mgd.$$
$$\frac{1}{2} kx^{2}= \mu_{k}mgd$$
$$\mu_{k}= \frac{kx^{2}}{2mgd} = \frac{(105 \; \text{N/m})(-0.100 \; \text{m}^{2})}{2(2.00 \; \text{kg})(9.81 \; \text{m/s}^{2}) (0.250 \; \text{m})}= \boxed{0.107}.$$

So if I analyze this solution, I guess the step that pops out at me immediately is this: how can they know that $-KE_{i} = -PE_{e}$ (assuming it was a typo above, as I already mentioned)?
It looks a bit convoluted actually.
Seems the solution is using conservation of energy without explicitly stating that it does.

This solution does not look entirely valid.
It refers to $KE_{f}$ and $KE_{i}$, but those are both zero.
It seems the KE in state 2 is intended, but if it is, then part of the friction is already overlooked (where the spring is decompressing).
So it seems the spring decompresses with no friction on the block, and it's only afterward that the distance is counted.
As I see it, KE should not be a part of the solution.

Furthermore, the formula for $W_{net}$ contains a typo.
The minus sign should be an equal sign.

#### Ackbach

##### Indicium Physicus
Staff member
It looks a bit convoluted actually.
Seems the solution is using conservation of energy without explicitly stating that it does.
Right, exactly.

This solution does not look entirely valid.
I agree.

It refers to $KE_{f}$ and $KE_{i}$, but those are both zero.
It seems the KE in state 2 is intended,
Doubtless.

but if it is, then part of the friction is already overlooked (where the spring is decompressing).
Again, I quite agree.

So it seems the spring decompresses with no friction on the block, and it's only afterward that the distance is counted.
Yes, that is what it seems. What's odd, though, is that they get the same thing as I do above using what I'm pretty sure is a valid calculus-based approach. Might be serendipitous.

As I see it, KE should not be a part of the solution.
You're saying the #2 event (where the mass reaches the natural length of the spring) is not necessary to solve the problem?

Furthermore, the formula for $W_{net}$ contains a typo.
The minus sign should be an equal sign.
Hmm. It's definitely a minus sign in the book's solution. So $\cos(\theta)=1$? But isn't the force of friction opposed to the direction of motion? Then shouldn't $\cos(\theta)=-1$?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, that is what it seems. What's odd, though, is that they get the same thing as I do above using what I'm pretty sure is a valid calculus-based approach. Might be serendipitous.
It fits.
The result would be the same.
You can either count the distance from the compressed state and include the decompression in the friction (which you should).
Or you can first decompress without friction, and then count the distance from the decompressed state.
In both cases the amount of work by friction (or rather the distance) is the same.
The latter case is convoluted though.

You're saying the #2 event (where the mass reaches the natural length of the spring) is not necessary to solve the problem?
Yes.
It only distracts, since you have to split the problem in 2 parts, which is not necessary.

Hmm. It's definitely a minus sign in the book's solution. So $\cos(\theta)=1$? But isn't the force of friction opposed to the direction of motion? Then shouldn't $\cos(\theta)=-1$?
The sign isn't really relevant.
You can pick any sign you like as long as you make sure the magnitude of the work by friction is the same as the magnitude of the elastic energy.

Either way, $d \cos \theta \ne d \cos \theta - d$.
Undoubtedly intended, is that $\theta = 0$, meaning $d \cos \theta = d$.

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