# Spring and block on an inclined plane

• king_harsh
If there were static friction that would figure into the equilibrium position. I believe...No, the static friction would not factor into the equilibrium position since the block is sliding without friction. It would only come into play if the block were to start moving in the opposite direction, in which case it would provide a restoring force.f

#### king_harsh

Moved from a technical forum, so homework template missing
A block of mass 0.2 kg which slides without friction on a θ = 30° incline is connected to the top of the incline by a mass-less spring of relaxed length of 23.75 cm and spring constant 80 N/m as shown in the following figure.

(a) How far from the top of the incline does the block stop?
(b) If the block is pulled slightly down the incline and released, what is the period of the ensuing oscillations? [g = 10ms-2 ].

Can anyone help me in solving this ?

What have you done so far about it? I think (a) is somehow strangely formulated. I would rephrase the question as

(a) At which location does the block stay at rest?

king_harsh and DrClaude
As per PhysicsForums rules for homework help, you will have to provide an attempt at a solution.

Perhaps try thinking about the KE+ IE = Emech and relating that to the system.

A block of mass 0.2 kg which slides without friction on a θ = 30° incline is connected to the top of the incline by a mass-less spring of relaxed length of 23.75 cm and spring constant 80 N/m as shown in the following figure. View attachment 270445
(a) How far from the top of the incline does the block stop?
(b) If the block is pulled slightly down the incline and released, what is the period of the ensuing oscillations? [g = 10ms-2 ].

Can anyone help me in solving this ?
It's 25cm away from the top of the incline.

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It's 25cm away from the top of the incline.
The thread is two years old, which makes posting a full solution ok.

Unfortunately the question is not quite clear. You have found the equilibrium position, but it asks for where the block will "stop". So possibly we are supposed to assume it is released with the spring at its relaxed length and are to find where it first comes to a stop.

I think that it is more likely that in part (a) we are to assume that the mass comes to and remains at rest. That's because of part (b) which states that "if the block is pulled slightly down the incline and released, what is the period of the ensuing oscillations?" It is easier to pull the mass "slightly down the incline" when it is at rest than when it is moving. Also, if the mass were moving already in part (a), part (b) would just ask for the period of oscillations without needing to mention the initial displacement. That said, I agree that the question is not quite clear.

we are to assume that the mass comes to and remains at rest
Not comes to, it is frictionless. It could have been placed at the equilibrium position.

Not comes to, it is frictionless. It could have been placed at the equilibrium position.
I was thinking of a disembodied hand easing it gently in position. The same hand that pulls it down slightly in part (b).

SammyS
The thread is two years old, which makes posting a full solution ok.

Unfortunately the question is not quite clear. You have found the equilibrium position, but it asks for where the block will "stop". So possibly we are supposed to assume it is released with the spring at its relaxed length and are to find where it first comes to a stop.
If it slides without friction, it never comes to a stop. Unless instantaneously having zero velocity counts as "coming to a stop"?

Incidentally, if this was released while the spring was relaxed with friction. Is this the EOM?

$${}_{+}\swarrow \sum F = m \ddot x = mg \sin \theta - kx - \mu_k mg \cos \theta \frac{ \dot x}{| \dot x |}$$

How do we solve that problem to find where it comes to a stop. I've never encountered it before. Is there an analytical result?

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Are you considering a separate problem in which friction is present? In this problem there is no friction.

Unless instantaneously having zero velocity counts as "coming to a stop"?
I agree it is unclear, but that is a valid case.

erobz
Are you considering a separate problem in which friction is present? In this problem there is no friction.
Since this was old, and unclearly stated I was just asking a "what if", to expand to something that definitely would come to a stop.

Incidentally, if this was released while the spring was relaxed with friction. Is this the EOM?

$$m \ddot x = mg \sin \theta - kx - \mu_k mg \cos \theta \frac{ \dot x}{| \dot x |}$$

How do we solve that problem to find where it comes to a stop. I've never encountered it before. Is there an analytical result?
If there were static friction that would figure into the equilibrium position. I believe the kinetic friction would just get one there faster.

If there were static friction that would figure into the equilibrium position. I believe the kinetic friction would just get one there faster.
I don't think I'm following along. If there is motion, it is kinetic friction dissipation the energy. Are you saying there is no "final" position for the mass in what I proposed above. The oscillations just exponentially decay, but never cease?

If box is sliding along on a flat surface, it eventually comes to a stop. Kinetic friction stopped it, not static friction.

?

Since this was old, and unclearly stated I was just asking a "what if", to expand to something that definitely would come to a stop.
I considered a simpler problem where we have a spring-mass system on a horizontal rough plane. Then the equation reduces to $$m\ddot x=-kx-\mu mg\frac{\dot x}{|\dot x|}.$$ Suppose we start the mass with initial velocity ##v_0## from the relaxed position of the spring and want to find the position of the first occurrence of zero velocity. This is a standard intro physics problem that involves solving a quadratic equation using the work-energy theorem. However, we will attempt a solution of th ODE as is. Since the friction force is constant, we can rewrite the ODE for the first trip out as $$m\ddot x=-k\left(x+\frac{f}{k}\right)~;~~(f \equiv \mu mg).$$With ##\omega^2=k/m##, the solution is $$x=\frac{v_0}{\omega}\sin\omega t-\frac{f}{\omega^2}.$$In principle, this is maximum when the sine is maximum, but this answer would be incorrect for the simple reason that this ##x(t)## implies ##\dot x(t)=v_0\cos \omega t##, which is the velocity without friction. It cannot be correct because the velocity decreases not only because the spring is puling the mass back but also because friction opposes the motion.

So, you see, the problem is with the approximation that the force of friction is independent of the velocity. When friction is constant, it just shifts the equilibrium position away from the relaxed length of the spring. This shift is just like when the plane on which the mass moves is frictionless and tilted at an angle with respect to the horizontal and a constant force ##mg\sin\theta## is added to the right hand side. When both friction and incline are present, the shift from the relaxed length will be a composite of the two. When the velocity goes past its first occurrence of zero, the contribution to the shift from friction will change sign but not that from gravity which is conservative. It looks like a mess.

I think the equation of motion for damped harmonic motion with a term proportional to the velocity would be more appropriate as a model.

Since the friction force is constant, we can rewrite the ODE for the first trip out as $$m\ddot x=-k\left(x+\frac{f}{k}\right)~;~~(f \equiv \mu mg).$$With ##\omega^2=k/m##, the solution is $$x=\frac{v_0}{\omega}\sin\omega t-\frac{f}{\omega^2}.$$

I get that:

$$x(t) = \frac{f}{k} \cos( \omega t) + \frac{v_o}{\omega} \sin ( \omega t) - \frac{f}{k }$$

So velocity is not ## \dot x(t) = v_o \cos ( \omega t) ##.

$$\dot x(t) = v_o \cos (\omega t) - \frac{\omega f }{k} \sin ( \omega t )$$

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Yes, you are correct. I made a silly mistake. I should not be doing this because it's late where I am and I'm tired so I will stop here. I guess the next step would be to find the position ##x_{\text{max}}## when ##\dot x(t)=0## and then see if it's the same as the value you get from the work-energy theorem. That should verify the consistency of the expressions. Finally, you might wish to see what lies beyond the first occurrence of zero velocity in which case you will have to set ##f\rightarrow -f## in the ODE and use initial conditions ##x(0)= x_{\text{max}}## and ##\dot x(0)=0.## I cannot think of doing this in another way than piecewise because that force of friction changes discontinuously at the time when the mass comes instantaneously to rest.

erobz
Yes, you are correct. I made a silly mistake. I should not be doing this because it's late where I am and I'm tired so I will stop here. I guess the next step would be to find the position ##x_{\text{max}}## when ##\dot x(t)=0## and then see if it's the same as the value you get from the work-energy theorem. That should verify the consistency of the expressions. Finally, you might wish to see what lies beyond the first occurrence of zero velocity in which case you will have to set ##f\rightarrow -f## in the ODE and use initial conditions ##x(0)= x_{\text{max}}## and ##\dot x(0)=0.## I cannot think of doing this in another way than piecewise because that force of friction changes discontinuously at the time when the mass comes instantaneously to rest.
How many times should I expect to repeat this process until it stops? I'm just curious if only ever approaches "stopped" in this model, never quite reaching it.

From work energy:

$$\mu_k m g \int dx = \frac{1}{2} m v_o^2 \implies \int dx = \frac{1}{2} \frac{v_o^2}{\mu_k g}$$

It seems like we can find the total displacement quite simply, but it's not so easy (possibly quite the understatement) to find where exactly it stops?

not so easy to find where exactly it stops.
I would think one could find the sequence of distances at each reversal.

I would think one could find the sequence of distances at each reversal.
So, just look for a pattern after doing a few iterations?

vanhees71
I also wonder if fundamentally there is a problem: How does it stop under the action of ##\mu_k##? I believe @bob012345 was heading in the same direction as @kuruman when he stated that the friction is probably better approximated by a velocity dependent force.

I'm imagining the oscillations exponetially decaying as the total displacement approches what I figured in post #20 under the force ##\mu_k mg ##, then in an "instant" the block stops with a force ## \mu_s mg > \mu_k mg ## acting on it, and the spring is either compressed or extended by some amount. This seems to make my head spin.

What I'm thinking is if we allow ##\mu_k## to "stop" the mass, then the energy stored in the spring (either in tension or compression) after any displacement of the mass ##D< \int dx ## must be tending to zero, and it necessarily stops when the spring is neither compressed or stretched... i.e. the free length of the spring seems to be the implication for the final resting place of the mass?

It seems like a bit of a Gabriels Horn Paradox, oscillations that never stop, yet have a finite displacement?

I would say draw the FBD at a turning point with the spring pulling in one direction and the force of static friction in the opposite direction. For simplicity, assume that the coefficient of static friction is equal to the coefficient of kinetic friction (it can be greater but not smaller). The mass will stop at the ##i##th turning point when the following condition is satisfied $$kx_{\text{max,i}}<\mu mg \implies x_{\text{max,i}}<\frac{\mu g}{\omega^2}.$$

erobz
I would say draw the FBD at a turning point with the spring pulling in one direction and the force of static friction in the opposite direction. For simplicity, assume that the coefficient of static friction is equal to the coefficient of kinetic friction (it can be greater but not smaller). The mass will stop at the ##i##th turning point when the following condition is satisfied $$kx_{\text{max,i}}<\mu mg \implies x_{\text{max,i}}<\frac{\mu g}{\omega^2}.$$
I feel like the model (as is) prohibits it from "stopping" at any location other than ##l_o##. Whenever it would "stop" by conservation of energy the amount of energy converted to heat by the friction force would be ##\frac{1}{2}m v_o^2##. If it stops at any location other than ##l_o## some amount of energy remains stored in the spring. I think we should find that the in limit the final "resting" place is ##l_o## in this model?

I feel like the model (as is) prohibits it from "stopping" at any location other than ##l_o##. Whenever it would "stop" by conservation of energy the amount of energy converted to heat by the friction force would be ##\frac{1}{2}m v_o^2##. If it stops at any location other than ##l_o## some amount of energy remains stored in the spring. I think we should find that the in limit the final "resting" place is ##l_o## in this model?
First, if I understand your problem, you are letting the block go with zero velocity from the unstretched spring position which implies it will never compress the spring, only stretch it. Then, if you are trying to model how the block will stop dynamically, I think that is outside the bounds of the simple model of friction we have with simple coefficients ## \mu_s## and ##\mu_k ##. These are experimentally determined coefficients according to a simple model but do not capture the dynamics of speed and the transition from kinetic friction to static friction. The best thing to do is model where the static equilibrium is and know that is where it will end up in the end. Then model the oscillations with ##\mu_k ## and define when it is stopped according to some threshold where the mathematical oscillations are small.

erobz
First, if I understand your problem, you are letting the block go with zero velocity from the unstretched spring position which implies it will never compress the spring, only stretch it.
No, we were talking about the (geometrically) simplified version @kuruman proposed to help explain the initial equation. The block is on a horizontal surface, it has some initial velocity ##v_o## at ##x=0##. It stretches and compresses.

The rest of what you say I agree with. The model poorly captures reality. In reality the block comes to rest in with some energy stored in the spring., like @kuruman alludes to in #25. I was just saying that using this simplified model seems to suggest that, if we were to find the limit using the piecewise ODE solutions and resulting sequence, the final position should be ##l_o## for this overly simplistic model.

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