Variable friction on an inclined plane and maximum velocity

[rockinwhiz]

Homework Statement

A small block when released on an incline plane, first slides down and then stops after sliding down a height $h$. This strange behaviour is due to the coefficient of friction that is here proportional to the distance slid by the block. Find the maximum speed of the block during this motion. Acceleration due to gravity is $g$.

Relevant Equations

$$U=mgh$$
$$F_s= \mu N$$

This problem was from the chapter on Work and Energy so, I thought of using the principle of conservation of mechanical energy. Clearly, the potential energy of the block decreases by mgh (assuming the block has mass m). This energy should have been converted to kinetic energy, but it clearly hasn't. That means work has been done on it by the friction of the inclined plane. However, I don't know how to proceed from here as I can't figure out how work by friction enters the equation. Only thing I can say is that μ=kx where μ is the coefficient of sliding friction, k is a constant and x is the distance slid by the block.
Kindly help me.
Thank you.

 

[TSny]

Are you familiar with the work-energy theorem which relates the total work done to the kinetic energy?

 

[rockinwhiz]

Are you familiar with the work-energy theorem which relates the total work done to the kinetic energy?

Is that the work energy theorem? That change in kinetic energy is the work done? Then yes sir, I know that.

 

[TSny]

Is that the work energy theorem? That change in kinetic energy is the work done? Then yes sir, I know that.

Yes. Can you find the total work done when the block moves a distance x?

 

[rockinwhiz]

Yes. Can you find the total work done when the block moves a distance x?

Using kinematic equations and the work energy theorem?

 

[TSny]

Use the definition of work to find the work done by each force acting on the block as the block moves a distance x.

 

[rockinwhiz]

Use the definition of work to find the work done by each force acting on the block as the block moves a distance x.

This is actually the part where I'm facing problems. I know the work sums to 0 but I don't know how to handle the work done by the frictional force, especially since it is variable.

 

[madafo3435]

You should be aware that friction is not a conservative force. help yourself with this expression. If you want, I can share the book from where I got it and in which this topic is discussed very well.

sorry if my english is bad, my source language is spanish

 

[rockinwhiz]

You should be aware that friction is not a conservative force. help yourself with this expression. If you want, I can share the book from where I got it and in which this topic is discussed very well.

sorry if my english is bad, my source language is spanish

Please do tell which book this is from. I believe the W stands for work done by non-conservative forces? Well, here's what, in the problem, the change in potential energy is evident, there's no change in kinetic energy, so work done by friction is apparently $mgh$, but how do I proceed from there?

 

[TSny]

This is actually the part where I'm facing problems. I know the work sums to 0 but I don't know how to handle the work done by the frictional force, especially since it is variable.

The work will not sum to zero for a general distance x. The work-energy theorem says that the work will sum to the change in KE.

Identify all of the forces acting on the block. Find an expression for the work done by each force after the block has moved a distance x. Express the work in terms of x.

As a start, pick anyone of the forces acting on the block and tell us what you get for the work done by that force.

 

[rockinwhiz]

The work will not sum to zero for a general distance x. The work-energy theorem says that the work will sum to the change in KE.

Identify all of the forces acting on the block. Find an expression for the work done by each force after the block has moved a distance x. Express the work in terms of x.

As a start, pick anyone of the forces acting on the block and tell us what you get for the work done by that force.

Okay, that's a very good point that I overlooked. So, as you say, if I pick the gravitational force and pick an arbitrary height $x$ that it descends through, then $mgx=(mv^2)/2$ isn't it (as initially the kinetic energy is zero)? Does that look right? But again, I have my doubts with the frictional force...

 

[madafo3435]

Please do tell which book this is from. I believe the W stands for work done by non-conservative forces? Well, here's what, in the problem, the change in potential energy is evident, there's no change in kinetic energy, so work done by friction is apparently $mgh$, but how do I proceed from there?

the boock is AN INTRODUCTION TO MECHANICS BY DANIEL KLEPPNER, second edition pg 188.
in effect, W represents work by friction, you can evaluate this quantity by making the line integral of this force. If you want, I can proceed with trying a solution

 

[rockinwhiz]

BTW I appreciate all this help a lot. Thank you. This'll probably get me better insight into the topic. Thanks again.

 

[rockinwhiz]

the boock is AN INTRODUCTION TO MECHANICS BY DANIEL KLEPPNER, second edition pg 188.
in effect, W represents work by friction, you can evaluate this quantity by making the line integral of this force. If you want, I can proceed with trying a solution

Please do try out the solution, but if you could provide just some guidelines and hints, I think that'd help me even more, as I'd understand things better. Thanks.

 

[TSny]

Okay, that's a very good point that I overlooked. So, as you say, if I pick the gravitational force and pick an arbitrary height x that it descends through, then mgx=(mv2)/2 isn't it (as initially the kinetic energy is zero)? Does that look right? But again, I have my doubts with the frictional force...

Initially, you used x for the distance the block slides along the incline. So, it will get confusing if you also use x for the height that the block descends. Maybe use h for the vertical height that the block descends. Using trig you can relate h to x.

The force of gravity is not the only force that does work on the block. So, it is not correct to set the work done by gravity equal to the change in kinetic energy. It's the total work done by all of the forces that equals the change in kinetic energy. That's the statement of the work-energy theorem. That's why you need to identify all of the forces acting on the block and find the work done separately by each force. Then you can sum the works and set the total work equal to the change in kinetic energy.

The formula given by @madafo3435 might appear to be something different. But, it's just a different way of expressing the work-energy theorem. The work done by the force of gravity can be expressed in terms of change in gravitational potential energy. So, you have a choice in dealing with the force of gravity. You can use the formula of madafo3435 in which the potential energy terms take care of the work done by gravity, or you can use the work-energy theorem in the form $\Delta KE = \sum W$ and include the work done by gravity on the right side.

 

[madafo3435]

I agree with @TSny, you must analyze well what forces exert work, start by analyzing normal force, it does not exert work so it is not considered in the work-energy relationship (can you see why?). Then, I suggest the following, draw a relationship between the path and the angle of the inclined plane, consider after 3 moments, when part of rest, when it returns to rest and when it reaches maximum speed, this last moment can be found when acceleration is 0 (why?), then combine these results with the work-energy relationship. I also suggest that you detail the reference system you are taking

 

[rockinwhiz]

Okay, here's my attempt at trying out all that you've suggested:
First off, let $\theta$ be the angle between the inclined plane and the vertical. Then aligning the axes along the inclined planed(X along the plane and Y perpendicular to it), we have the force $mg \cos \theta$ along the positive X-axis, $f$ the frictional force along the negative X-axis. Then the weight of the body acting on the inclined plane is $mg \sin \theta$ which is equal to the normal force $N$. Now $f= \mu N$. If the body slides down a distance $x$, work done by gravity on it is $mgx \cos \theta$ and work by friction would be $\int_0^x \! f \, \mathrm{d}x$(integral as it is non-constant) . This sums to the change in kinetic energy, which is simply $mv^2/2$ as initial velocity is 0.

Now, as madafo3435 said, the acceleration would be zero at maximum Velocity(derivative is zero at this point as curve will be at maxima) so I suppose $f$ and $mg \cos \theta$ balance each other out. This is all I can think of so far.

 

[rockinwhiz]

BTW my late reply is probably due to difference in time zones.

 

[haruspex]

This sums to the change in kinetic energy

So substitute for f using its known form and perform the integral.

 

[madafo3435]

Okay, here's my attempt at trying out all that you've suggested:
First off, let $\theta$ be the angle between the inclined plane and the vertical. Then aligning the axes along the inclined planed(X along the plane and Y perpendicular to it), we have the force $mg \cos \theta$ along the positive X-axis, $f$ the frictional force along the negative X-axis. Then the weight of the body acting on the inclined plane is $mg \sin \theta$ which is equal to the normal force $N$. Now $f= \mu N$. If the body slides down a distance $x$, work done by gravity on it is $mgx \cos \theta$ and work by friction would be $\int_0^x \! f \, \mathrm{d}x$(integral as it is non-constant) . This sums to the change in kinetic energy, which is simply $mv^2/2$ as initial velocity is 0.

Now, as madafo3435 said, the acceleration would be zero at maximum Velocity(derivative is zero at this point as curve will be at maxima) so I suppose $f$ and $mg \cos \theta$ balance each other out. This is all I can think of so far.

I agree with your reasoning, it's the easiest way I can find to include speed

 

[rockinwhiz]

Okay, here I go:
$1)$ At maximum velocity, acceleration is zero. So, $f-mg \cos \theta = 0$. This implies: $\mu N \equiv kxmg \sin \theta= mg \cos \theta$. So, $k= \cot \theta /x$.

$2)$ Now, since Total Work= Change in Kinetic Energy, hence, $mgx \cos \theta + \int_{0}^{x} f dx = mv^2/2$. Substituting the value of $k$, as obtained in step 1 into $f$ gives $f= mg \cos \theta$. Thus, $mgx \cos \theta + \int_{0}^{x} mg \cos \theta dx = mv^2/2$ which implies $2mgx \cos \theta=mv^2/2$ or, $v= \sqrt {4 gx \cos \theta}$. Now, $x \cos \theta$ is simply the height through which the block has descended. The only question left now is: how do I maximize this?

 

[haruspex]

gives $f= mg \cos \theta$.

You are forgetting $\mu$, which is a function of x.

 

[rockinwhiz]

You are forgetting $\mu$, which is a function of x.

I'm totally lost now sir. Could you kindly give an outline of the required equation?

 

[haruspex]

I'm totally lost now sir. Could you kindly give an outline of the required equation?

You have $N=mg\sin(\theta)$, $f=\mu N$, $\mu=kx$, $W=\int_0^d f.dx$.
Put those together to find W as a function of m, g, $\theta$, k and d.
You also need an equation relating $\theta$ and d to h.

 

[madafo3435]

Okay, here I go:
$1)$ At maximum velocity, acceleration is zero. So, $f-mg \cos \theta = 0$. This implies: $\mu N \equiv kxmg \sin \theta= mg \cos \theta$. So, $k= \cot \theta /x$.

$2)$ Now, since Total Work= Change in Kinetic Energy, hence, $mgx \cos \theta + \int_{0}^{x} f dx = mv^2/2$. Substituting the value of $k$, as obtained in step 1 into $f$ gives $f= mg \cos \theta$. Thus, $mgx \cos \theta + \int_{0}^{x} mg \cos \theta dx = mv^2/2$ which implies $2mgx \cos \theta=mv^2/2$ or, $v= \sqrt {4 gx \cos \theta}$. Now, $x \cos \theta$ is simply the height through which the block has descended. The only question left now is: how do I maximize this?

If you allow me, we must add that the x you are using in step 2) is the distance at which the maximum speed is given. Now, I am not sure but I think that by analyzing the work-energy relationship from this point until it returns to rest, we will obtain more information that allows us to eliminate x from its relation, in short, I suggest that you repeat the procedure from step 2) on the missing path.
What do you think?

 

[rockinwhiz]

If you allow me, we must add that the x you are using in step 2) is the distance at which the maximum speed is given. Now, I am not sure but I think that by analyzing the work-energy relationship from this point until it returns to rest, we will obtain more information that allows us to eliminate x from its relation, in short, I suggest that you repeat the procedure from step 2) on the missing path.
What do you think?

You have $N=mg\sin(\theta)$, $f=\mu N$, $\mu=kx$, $W=\int_0^d f.dx$.
Put those together to find W as a function of m, g, $\theta$, k and d.
You also need an equation relating $\theta$ and d to h.

Okay, let me try. I'll reply soon.

 

[rockinwhiz]

I think I've got it. First off, the total height is $h$. Let the incline's length be $d$ and the angle between it and the vertical $\theta$. Clearly, $d \cos \theta =h$. Firstly, the acceleration on the particle $= g \cos \theta -kgx \sin \theta$. Now here, $a=dv/dt=(dv/dx) \cdot v$. Cross multiplying both sides with $dx$ and integrating from $0$ to $d$ gives $0=gd \cos \theta -(kd^2g \sin \theta) /2$ (LHS is 0 as velocity is 0 at the beginning point and at $d$). So, after some simple algebra $d = 2 \cot \theta /k= h /( \cos \theta)$, yielding $h=2 \cos \theta \cot \theta /k$. Then, adopting the same method over distance $l$, where $v$ is maximum, I get $v^2/2=gl \cos \theta + (kl^2 g \sin \theta)/2$. Now, differentiating both sides with respect to $l$ gives $0=g \cos \theta - klg \sin \theta$(v is max here, so its derivative is zero). I end up with $k \cot \theta = l$. Putting $l$'s value in the last equation, simplifying, finally substituting $k$'s value gives me:
$$v= \sqrt { \frac {gh} 2} $$

There might be some errors here and there because I've typed it VERY briefly and hastily, but I think this is it. Because, the given answer matches as well.

 

[rockinwhiz]

So, basically, I was missing out on Newton's Laws as well as some work-energy related points.

 

[haruspex]

I think I've got it. First off, the total height is $h$. Let the incline's length be $d$ and the angle between it and the vertical $\theta$. Clearly, $d \cos \theta =h$. Firstly, the acceleration on the particle $= g \cos \theta -kgx \sin \theta$. Now here, $a=dv/dt=(dv/dx) \cdot v$. Cross multiplying both sides with $dx$ and integrating from $0$ to $d$ gives $0=gd \cos \theta -(kd^2g \sin \theta) /2$ (LHS is 0 as velocity is 0 at the beginning point and at $d$). So, after some simple algebra $d = 2 \cot \theta /k= h /( \cos \theta)$, yielding $h=2 \cos \theta \cot \theta /k$. Then, adopting the same method over distance $l$, where $v$ is maximum, I get $v^2/2=gl \cos \theta + (kl^2 g \sin \theta)/2$. Now, differentiating both sides with respect to $l$ gives $0=g \cos \theta - klg \sin \theta$(v is max here, so its derivative is zero). I end up with $k \cot \theta = l$. Putting $l$'s value in the last equation, simplifying, finally substituting $k$'s value gives me:
$$v= \sqrt { \frac {gh} 2} $$

There might be some errors here and there because I've typed it VERY briefly and hastily, but I think this is it. Because, the given answer matches as well.

Looks good.

 

[rockinwhiz]

A big thank you to all of you helping me out.

 

[madafo3435]

I think I've got it. First off, the total height is $h$. Let the incline's length be $d$ and the angle between it and the vertical $\theta$. Clearly, $d \cos \theta =h$. Firstly, the acceleration on the particle $= g \cos \theta -kgx \sin \theta$. Now here, $a=dv/dt=(dv/dx) \cdot v$. Cross multiplying both sides with $dx$ and integrating from $0$ to $d$ gives $0=gd \cos \theta -(kd^2g \sin \theta) /2$ (LHS is 0 as velocity is 0 at the beginning point and at $d$). So, after some simple algebra $d = 2 \cot \theta /k= h /( \cos \theta)$, yielding $h=2 \cos \theta \cot \theta /k$. Then, adopting the same method over distance $l$, where $v$ is maximum, I get $v^2/2=gl \cos \theta + (kl^2 g \sin \theta)/2$. Now, differentiating both sides with respect to $l$ gives $0=g \cos \theta - klg \sin \theta$(v is max here, so its derivative is zero). I end up with $k \cot \theta = l$. Putting $l$'s value in the last equation, simplifying, finally substituting $k$'s value gives me:
$$v= \sqrt { \frac {gh} 2} $$

There might be some errors here and there because I've typed it VERY briefly and hastily, but I think this is it. Because, the given answer matches as well.

In good time, his reasoning is very reasonable, but I must clarify that the signing force is not equal to -kmgxsen ($\theta$), it is actually -kmgx. Do you agree with me? Still, considering friction in this way, you may notice that you get the same result.

 

[rockinwhiz]

In good time, his reasoning is very reasonable, but I must clarify that the signing force is not equal to -kmgxsen ($\theta$), it is actually -kmgx. Do you agree with me? Still, considering friction in this way, you may notice that you get the same result.

Oh yeah. I guess I'll check it out later. Thanks for everything man.

 

[madafo3435]

Oh yeah. I guess I'll check it out later. Thanks for everything man.

you're welcome, our discussion was interesting

 

[haruspex]

In good time, his reasoning is very reasonable, but I must clarify that the signing force is not equal to -kmgxsen ($\theta$), it is actually -kmgx. Do you agree with me? Still, considering friction in this way, you may notice that you get the same result.

$\theta$ and k are not specified in the problem statement, they are unknown constants created by the solver. So whether the friction coefficient is written as $kx$ or $kx\sin(\theta)$ is irrelevant, provided it is done consistently.