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ParthKohli

  • one year ago

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: \[\mu _ k = \cases{ x \quad \rm{for~0\le x \le 1} \\ 0 \quad \rm otherwise}\]

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  1. ParthKohli
    • one year ago
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    |dw:1433251163959:dw|

  2. ParthKohli
    • one year ago
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    |dw:1433251254232:dw|

  3. ParthKohli
    • one year ago
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    @ganeshie8

  4. ParthKohli
    • one year ago
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    \[dW_f = f \cdot dx = \mu_k \cdot m_{\rm eff}g \cdot dx\]Now \(\mu_k = x\) and \(m_{\rm eff} = m - m\cdot \frac{x}{1} \)

  5. ganeshie8
    • one year ago
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    you want to calculate the work done by the friction as the block moves completely to the smooth surface

  6. ParthKohli
    • one year ago
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    Yes. Can it be done without double-integration?

  7. ParthKohli
    • one year ago
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    \[dW_{f} = x\cdot \left(m - mx\right)\cdot g\cdot dx\]Is that it?

  8. ganeshie8
    • one year ago
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    only one dimension is involved here, so a single integral will do right ?

  9. ParthKohli
    • one year ago
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    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.

  10. ParthKohli
    • one year ago
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    Just looking for your confirmation.

  11. ganeshie8
    • one year ago
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    that is where integral comes in if the friction is same everywhere then the work done is simply \(F*s\) right

  12. ParthKohli
    • one year ago
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    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. :)

  13. ganeshie8
    • one year ago
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    since the friction is changing, we need to setup an integral : \[\int_0^1 F\cdot ds\]

  14. ganeshie8
    • one year ago
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    I don't remember the exact friction formulas tho

  15. ParthKohli
    • one year ago
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    \[f_k = \mu_k \times N \]Here, \(f_k = \mu_k \times m g\) where \(m \) is the mass on which friction acts.

  16. ganeshie8
    • one year ago
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    \[w=\int_0^1 xmg~dx\] ?

  17. ParthKohli
    • one year ago
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    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?

  18. ganeshie8
    • one year ago
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    oh right not entire block is above friction surface as it moves

  19. ParthKohli
    • one year ago
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    So\[W_{net} = mg\int_{0}^1 (x - x^2)dx = mg \left[\frac{x^2}{2} - \dfrac{x^3}{3}\right]_{x = 0}^{x=1}\]\[= mg \left[\frac{1}{2} - \frac{1}{3}\right] = \frac{mg}{6}\]

  20. ParthKohli
    • one year ago
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    That isn't correct. :|

  21. ganeshie8
    • one year ago
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    whats the correct answer

  22. ganeshie8
    • one year ago
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    we did not factor in the horizontal force F in our calculation

  23. ParthKohli
    • one year ago
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    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 \(\frac{4}{3}\).

  24. ParthKohli
    • one year ago
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    Oh yes, but is it relevant?

  25. ParthKohli
    • one year ago
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    Isn't the work done by friction independent of the magnitude of force anyway?

  26. ganeshie8
    • one year ago
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    Oh right, we don't need net force on block, just the friction

  27. ParthKohli
    • one year ago
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    So our answer is coming out to be\[\frac{2 \cdot 10}{6} J= \frac{10}{3}J\]

  28. ganeshie8
    • one year ago
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    don't trust me on these things, but wait, what is the numerical value of given F

  29. ParthKohli
    • one year ago
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    No numerical value is given. :P

  30. ganeshie8
    • one year ago
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    oh then it shouldnt matter as you said haha

  31. ParthKohli
    • one year ago
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    Tagging @Kainui to check our work.

  32. ParthKohli
    • one year ago
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    The friction at each point is also different though.|dw:1433253462744:dw|

  33. ParthKohli
    • one year ago
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    We have done the integral assuming that friction remains the same at A and B and all other such points, right?

  34. ParthKohli
    • one year ago
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    Again, I'm not sure. The integral may compensate for that, but still, I think that's our catch.

  35. ParthKohli
    • one year ago
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    |dw:1433253743704:dw|

  36. ParthKohli
    • one year ago
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    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:\[W_{net, f}=g \int_{0}^{1} \int_0^{M(1-x)} xdmdx\]

  37. ganeshie8
    • one year ago
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    do you have 20/3 in options

  38. ParthKohli
    • one year ago
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    Yes - how did you get that?

  39. ganeshie8
    • one year ago
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    The trick is to call "x" as the length of block that is on the friction surface

  40. Kainui
    • one year ago
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    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.

  41. ParthKohli
    • one year ago
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    @Kainui You're actually right. You just forgot the factor of \(g\).

  42. ganeshie8
    • one year ago
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    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 \[\int\limits_0^1 \mu_k*mx*g~dx\]

  43. ParthKohli
    • one year ago
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    Oh, wow, that sounds easier. Where did our setup go wrong?

  44. ganeshie8
    • one year ago
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    we assumed "x" as a point on surface instead of varying length of the block on the friction surface

  45. ParthKohli
    • one year ago
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    Oh...

  46. ganeshie8
    • one year ago
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    setting up correct integral is somewhat pain if we treat x as a point on the surface

  47. ParthKohli
    • one year ago
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    I see. The difference in our integrals is that \(\mu_k = x\) gives the correct answer in your case.

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