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Consider a body attached to a horizontal spring and resting on a surface, inclined at an angle $\theta$ from the ground.Set-up

The spring constant is $k$. Initially the spring was kept in its natural length while the body was held still by some external agent. When the external agent was removed, the body slid $x$ units down the inclined plane to achieve equilibrium.The coeffecient of (kinetic) frictional force acting on the body is $\mu$

To solve this problem, I can use two methods:

1.Work-Energy Theorem: This yields $$0=mg(x\sin\theta)-\frac12kx^2-\mu (mgx\cos\theta)$$

2.Equating forces at equilibrium: This yields $$kx+\mu mg\cos\theta-mg\sin\theta=0$$(along the inclined plane)

On solving the equations the two methods give different answers, while they should give the same. Is there anything I am missing out here? Please help me solve this problem.

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  • $\begingroup$ What is the actual question you're trying to solve? Are you solving for $x$? $\endgroup$
    – BMS
    Commented Sep 10, 2014 at 16:55
  • $\begingroup$ Yes. However, as you can see, I cannot seem to derive a consistent solution $\endgroup$
    – user117913
    Commented Sep 10, 2014 at 17:00
  • $\begingroup$ Can you add that piece of information, and any other relevant factors, to your question? Also, that means your title may need changing since you aren't solving for $\mu$. $\endgroup$
    – BMS
    Commented Sep 10, 2014 at 17:03
  • $\begingroup$ Note that your equations are exactly equivalent except for the factor $\frac{1}{2}$ in front of the spring-energy term. $\endgroup$
    – Danu
    Commented Sep 10, 2014 at 17:05
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    $\begingroup$ $\mu m g \cos \theta$ is the maximum force of friction. At rest, this value may not be maximum. In fact, it may be negative. It may be acting downward to oppose a net upward force because the spring is pulling harder than gravity. So your null force equation appears to be too simplistic. I believe there will be a range of $x$ where the net force can be zero due to the variable amount of friction force. $\endgroup$
    – BowlOfRed
    Commented Sep 10, 2014 at 17:46

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@soumyadeep is on the right track, but wrong.

The object starts at rest. When the "external force" is removed, it starts to slide down the slope, picking up speed. When it reaches the point where the force of friction plus the force of the spring equal the force of gravity, it doesn't just stop - it stops accelerating.

Thus, to get equality you need to consider the net amount of energy that was available before the block reaches the extension $x$ - you will find that it's exactly the $\frac12 kx^2$ that you were missing. This is kinetic energy in the system at the time that the forces are in equilibrium.

Alternatively - if you wanted to make the block move very, very slowly, you would have to hold it back (compensate for the lack of force that the spring initially has) with a force that will change with displacement $x$ - the sum of this force and the force of the spring being always constant. If other words,

$$F_{ext} = k(L-x)$$

where $L$ is the full length where equilibrium occurs.

You can immediately see that the amount of work done by the block against this external force will be equal to the amount of energy stored in the spring at full extension ($x=L$) by integrating:

$$\begin{align}U &= \int_0^L k(L-x)dx\\ &= kL^2 - \frac12 kL^2\\ &= \frac12 kL^2\\ \end{align}$$

so once again you find your missing factor $\frac12 k L^2$ in the energy equation.

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  • $\begingroup$ Well,i have got it but can you should we not consider the thing BowlOfRed has commented about friction? $\endgroup$
    – soumyadeep
    Commented Sep 11, 2014 at 4:25
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    $\begingroup$ I don't think so because kinetic friction is assumed to be constant here. $\endgroup$
    – user117913
    Commented Sep 11, 2014 at 9:22
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    $\begingroup$ While @bowlofred's comment is in general true, it is assumed irrelevant in a question like this - with the above explanation the two methods give the same answer without invoking some unspecified behavior. So while it's true in a physics sense that friction is not constant, I don't need it to explain why your analyses gave different results. $\endgroup$
    – Floris
    Commented Sep 11, 2014 at 12:01

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