ParthKohli
  • ParthKohli
A block of mass \(m~kg\) and length \(1~m\) is placed on a horizontal rough surface having a variable kinetic friction coefficient. A horizontal force \(F\) is applied such that the block moves slowly. What is the total heat released in the process? The coefficient of friction varies as follows: \[\mu _ k = \cases{ x \quad \rm{for~0\le x \le 1} \\ 0 \quad \rm otherwise}\]
Mathematics
schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
|dw:1433251163959:dw|
ParthKohli
  • ParthKohli
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ParthKohli
  • ParthKohli

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ParthKohli
  • ParthKohli
\[dW_f = f \cdot dx = \mu_k \cdot m_{\rm eff}g \cdot dx\]Now \(\mu_k = x\) and \(m_{\rm eff} = m - m\cdot \frac{x}{1} \)
ganeshie8
  • ganeshie8
you want to calculate the work done by the friction as the block moves completely to the smooth surface
ParthKohli
  • ParthKohli
Yes. Can it be done without double-integration?
ParthKohli
  • ParthKohli
\[dW_{f} = x\cdot \left(m - mx\right)\cdot g\cdot dx\]Is that it?
ganeshie8
  • ganeshie8
only one dimension is involved here, so a single integral will do right ?
ParthKohli
  • ParthKohli
Hmm, yes. I think that is it. The reason why I was confused was because at each point, friction is different. I am not a smart man, I guess.
ParthKohli
  • ParthKohli
Just looking for your confirmation.
ganeshie8
  • ganeshie8
that is where integral comes in if the friction is same everywhere then the work done is simply \(F*s\) right
ParthKohli
  • ParthKohli
Yes, I understood now. I thought we needed to get a \(dm\) too, but turns out that \(m\) can be directly expressed as a function of \(x\). Thanks. :)
ganeshie8
  • ganeshie8
since the friction is changing, we need to setup an integral : \[\int_0^1 F\cdot ds\]
ganeshie8
  • ganeshie8
I don't remember the exact friction formulas tho
ParthKohli
  • ParthKohli
\[f_k = \mu_k \times N \]Here, \(f_k = \mu_k \times m g\) where \(m \) is the mass on which friction acts.
ganeshie8
  • ganeshie8
\[w=\int_0^1 xmg~dx\] ?
ParthKohli
  • ParthKohli
So if the block has been pushed \(x\) distance ahead and is pushed a further \(dx\), then the mass on which friction is acting is \(m(1 - x)\) right?
ganeshie8
  • ganeshie8
oh right not entire block is above friction surface as it moves
ParthKohli
  • ParthKohli
So\[W_{net} = mg\int_{0}^1 (x - x^2)dx = mg \left[\frac{x^2}{2} - \dfrac{x^3}{3}\right]_{x = 0}^{x=1}\]\[= mg \left[\frac{1}{2} - \frac{1}{3}\right] = \frac{mg}{6}\]
ParthKohli
  • ParthKohli
That isn't correct. :|
ganeshie8
  • ganeshie8
whats the correct answer
ganeshie8
  • ganeshie8
we did not factor in the horizontal force F in our calculation
ParthKohli
  • ParthKohli
It's in my assignment, and the correct answer isn't given. The original question says \(m = 2~kg\) and \( g = 10~ m/s^2\). The options are all multiples of \(\frac{4}{3}\).
ParthKohli
  • ParthKohli
Oh yes, but is it relevant?
ParthKohli
  • ParthKohli
Isn't the work done by friction independent of the magnitude of force anyway?
ganeshie8
  • ganeshie8
Oh right, we don't need net force on block, just the friction
ParthKohli
  • ParthKohli
So our answer is coming out to be\[\frac{2 \cdot 10}{6} J= \frac{10}{3}J\]
ganeshie8
  • ganeshie8
don't trust me on these things, but wait, what is the numerical value of given F
ParthKohli
  • ParthKohli
No numerical value is given. :P
ganeshie8
  • ganeshie8
oh then it shouldnt matter as you said haha
ParthKohli
  • ParthKohli
Tagging @Kainui to check our work.
ParthKohli
  • ParthKohli
The friction at each point is also different though.|dw:1433253462744:dw|
ParthKohli
  • ParthKohli
We have done the integral assuming that friction remains the same at A and B and all other such points, right?
ParthKohli
  • ParthKohli
Again, I'm not sure. The integral may compensate for that, but still, I think that's our catch.
ParthKohli
  • ParthKohli
|dw:1433253743704:dw|
ParthKohli
  • ParthKohli
Let's call the total mass \(M\) instead of \(m\). I'll use \(m\) as the index (is that what you call it?) If the block has been pushed from 0 to \(x\) and is pushed a further \(dx\), \(m\) varies from \(0\) to \(M(1-x)\).\[dW_f = \mu_k N \]\[= x\cdot dm\cdot g\cdot dx\]So net work:\[W_{net, f}=g \int_{0}^{1} \int_0^{M(1-x)} xdmdx\]
ganeshie8
  • ganeshie8
do you have 20/3 in options
ParthKohli
  • ParthKohli
Yes - how did you get that?
ganeshie8
  • ganeshie8
The trick is to call "x" as the length of block that is on the friction surface
Kainui
  • Kainui
Ok here's my attempt, hopefully it works out. The work done by an individual particle that starts at point s and slides to 1 will be: \[dW =\int_s^1 \bar f_k \cdot d \bar x = \int_s^1 xdm dx = \frac{1}{2}(1-s^2)dm\] In terms of linear density, I will rewrite dm as: \[dm = \gamma ds\] and we already know the linear density, so I'll write how to calculate it here:\[m \ kg=\gamma \ 1 m\] Now we in a place to add up all the paths of particles from 0 to 1 \[\int_0^1 dW = \int_0^1 \frac{1}{2}(1-s^2)m ds\] The left sides is just the total heat Q, and the right side integrates: \[Q = \frac{m}{3} \] But that's not a multiple of 4/3 so idk maybe I messed up my algebra somewhere along the way, I'm gonna make some breakfast and come back to see what's going on with this question.
ParthKohli
  • ParthKohli
@Kainui You're actually right. You just forgot the factor of \(g\).
ganeshie8
  • ganeshie8
Let \(x\) = the length of block on the friction surface, then the mass of block on the friction surface is given by \(mx\) work done in moving the block by "dx" units = \(dw = \mu_k *mx*g*dx\) work done in moving off the friction surface is \[\int\limits_0^1 \mu_k*mx*g~dx\]
ParthKohli
  • ParthKohli
Oh, wow, that sounds easier. Where did our setup go wrong?
ganeshie8
  • ganeshie8
we assumed "x" as a point on surface instead of varying length of the block on the friction surface
ParthKohli
  • ParthKohli
Oh...
ganeshie8
  • ganeshie8
setting up correct integral is somewhat pain if we treat x as a point on the surface
ParthKohli
  • ParthKohli
I see. The difference in our integrals is that \(\mu_k = x\) gives the correct answer in your case.

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