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

1. ParthKohli

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2. ParthKohli

|dw:1433251254232:dw|

3. ParthKohli

@ganeshie8

4. ParthKohli

$dW_f = f \cdot dx = \mu_k \cdot m_{\rm eff}g \cdot dx$Now $$\mu_k = x$$ and $$m_{\rm eff} = m - m\cdot \frac{x}{1}$$

5. ganeshie8

you want to calculate the work done by the friction as the block moves completely to the smooth surface

6. ParthKohli

Yes. Can it be done without double-integration?

7. ParthKohli

$dW_{f} = x\cdot \left(m - mx\right)\cdot g\cdot dx$Is that it?

8. ganeshie8

only one dimension is involved here, so a single integral will do right ?

9. 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.

10. ParthKohli

11. ganeshie8

that is where integral comes in if the friction is same everywhere then the work done is simply $$F*s$$ right

12. 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. :)

13. ganeshie8

since the friction is changing, we need to setup an integral : $\int_0^1 F\cdot ds$

14. ganeshie8

I don't remember the exact friction formulas tho

15. ParthKohli

$f_k = \mu_k \times N$Here, $$f_k = \mu_k \times m g$$ where $$m$$ is the mass on which friction acts.

16. ganeshie8

$w=\int_0^1 xmg~dx$ ?

17. 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?

18. ganeshie8

oh right not entire block is above friction surface as it moves

19. 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}$

20. ParthKohli

That isn't correct. :|

21. ganeshie8

22. ganeshie8

we did not factor in the horizontal force F in our calculation

23. 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 $$\frac{4}{3}$$.

24. ParthKohli

Oh yes, but is it relevant?

25. ParthKohli

Isn't the work done by friction independent of the magnitude of force anyway?

26. ganeshie8

Oh right, we don't need net force on block, just the friction

27. ParthKohli

So our answer is coming out to be$\frac{2 \cdot 10}{6} J= \frac{10}{3}J$

28. ganeshie8

don't trust me on these things, but wait, what is the numerical value of given F

29. ParthKohli

No numerical value is given. :P

30. ganeshie8

oh then it shouldnt matter as you said haha

31. ParthKohli

Tagging @Kainui to check our work.

32. ParthKohli

The friction at each point is also different though.|dw:1433253462744:dw|

33. ParthKohli

We have done the integral assuming that friction remains the same at A and B and all other such points, right?

34. ParthKohli

Again, I'm not sure. The integral may compensate for that, but still, I think that's our catch.

35. ParthKohli

|dw:1433253743704:dw|

36. 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$

37. ganeshie8

do you have 20/3 in options

38. ParthKohli

Yes - how did you get that?

39. ganeshie8

The trick is to call "x" as the length of block that is on the friction surface

40. 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.

41. ParthKohli

@Kainui You're actually right. You just forgot the factor of $$g$$.

42. 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 $\int\limits_0^1 \mu_k*mx*g~dx$

43. ParthKohli

Oh, wow, that sounds easier. Where did our setup go wrong?

44. ganeshie8

we assumed "x" as a point on surface instead of varying length of the block on the friction surface

45. ParthKohli

Oh...

46. ganeshie8

setting up correct integral is somewhat pain if we treat x as a point on the surface

47. ParthKohli

I see. The difference in our integrals is that $$\mu_k = x$$ gives the correct answer in your case.