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I was working on a modified version of this problem (which I solved):

enter image description here

The modification is as follows:

Consider the following system with two masses, $M$ and $m$, with the length of the body $M$ given to be $L$. The body $m$ is released from a height of $\frac{L}{2}$ along the ramp of $M$. The flatbed of length $L$ has a coefficient of friction $\mu$. Where on the bed, will the mass $m$ finally lie? enter image description here

I tried approaching the question by analysing how the relative velocity of $m$ w.r.t. $M$ changes while travelling on the bed. But, the relative velocity changes at every instant as $m$ travels over $M$. Also, just using the location of the center of mass to be constant doesn't help since neither of the bodies' locations is known in the final state. However, the acceleration of individual bodies is constant throughout the motion, and this fact could be helpful.

I, therefore, tried the following:

  • Acceleration of $m$, $\vec a_m = -\mu g \hat{i}$ (due to friction).
  • Acceleration of $M$, $\vec a_M = \mu g \frac{m}{M} \hat i$.

Thus, the acceleration of $m$ w.r.t. $M$ is $\mu mg\cdot(1+\frac{m}{M})$. Subsequently, the work done by friction over a distance $d$ shall equal the intitial gravitational potential energy of $m$:

\begin{align}\mu mg\cdot(1+\frac{m}{M})\cdot d = \frac{mgL}{2}\end{align}

The final position of $m$ will now be fairly easy to calculate after plugging in the numbers.

But I am not sure if what I have done is correct since according to the work equation above, it is as if all the "relative" frictional force works only on $m$, which in relative terms makes sense but the energy dissipation part doesn't. Shouldn't we also take into account $M$? I am not able to resolve this suspicion by myself. It would be great if I can get an explanation for whether my approach and the answer was correct or incorrect and why. Also, an alternative approach to this question will be highly appreciated.

Thanks!

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  • $\begingroup$ -1 Your explanation is very unclear. The explanation below your equation looks good to me, but I don't understand how you got it, and what is $d$? I don't understand what you say in the paragraph before it, nor what you write afterwards : ... here the work includes relative acceleration and all of the frictional force acts on $m$, which is not true. Friction is independent of relative acceleration, and all of the friction force does act on $m$. You need to explain more clearly what is bothering you. $\endgroup$ Mar 5, 2018 at 22:48
  • $\begingroup$ Hi! I'm sorry, I'll just edit it and make it more informative and clear. $\endgroup$ Mar 6, 2018 at 0:18
  • $\begingroup$ The work done against friction is friction force x distance. It has nothing to do with relative acceleration. Distance moved by block is measured relative to the bed, not the ground. For example, consider (1) the bed is stationary while the block moves (relative to the ground), (2) the block is stationary while the bed moves, (3) both block and bed move. If in each case the block starts at one end of the bed and stops at the other, then the distance moved against friction is the same in all 3 cases. But the distances relative to the ground are different. $\endgroup$ Mar 7, 2018 at 0:28
  • $\begingroup$ Yes, that makes sense. So will initial relative velocity squared divided relative acceleration give the correct answer? I don't know, but it seems wrong $\endgroup$ Mar 7, 2018 at 6:56
  • $\begingroup$ the work done by friction over a distance d shall equal the initial gravitational potential energy of m. That is all you need to know to solve the problem. There is no need to calculate any velocities or accelerations. Initial GPE is dissipated by friction between block and bed, which is the same in both versions of the problem. The final position of m relative to the bed is the same whether the bed moves or not. The final position relative to the ground is different, and can be found from the fact that in v2 the CM of the block+bed does not move; in v1 the CM of the bed does not move. $\endgroup$ Mar 7, 2018 at 12:50

1 Answer 1

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Both of these problems are about global energy conservation, not local forces, velocities, or accelerations. You have to dissipate the gravitational potential energy:

$$ E_0 = mgh = \frac 1 2 m g L .$$

Because of friction, it cost energy to traverse a distance (relative to the sled-thingy):

$$ \frac{dE}{dx}= \frac d {dx}W(x) $$

where the work is to go a distance $x$ is

$$ W(x) =\int_{t=0}^{t=x}{Fdt} = Fx $$

and the force is the frictional resistance:

$$ F = mg\mu$$

so that:

$$ \frac{dE}{dx}=mg\mu $$

The total distance, $d$, traveled is:

$$ E_0/(dE/dx) = \frac{\frac 1 2 m g L}{mg \mu}=\frac{L}{2\mu}.$$

As @sammy gerbil pointed out in the comments, this is relative to the bed, not the absolute position of the block.

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