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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: \[\mu _ k = \cases{ x \quad \rm{for~0\le x \le 1} \\ 0 \quad \rm otherwise}\]
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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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433251163959:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433251254232:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[dW_f = f \cdot dx = \mu_k \cdot m_{\rm eff}g \cdot dx\]Now \(\mu_k = x\) and \(m_{\rm eff} = m  m\cdot \frac{x}{1} \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you want to calculate the work done by the friction as the block moves completely to the smooth surface

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes. Can it be done without doubleintegration?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[dW_{f} = x\cdot \left(m  mx\right)\cdot g\cdot dx\]Is that it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1only one dimension is involved here, so a single integral will do right ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Hmm, 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
 one year ago
Best ResponseYou've already chosen the best response.2Just looking for your confirmation.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that is where integral comes in if the friction is same everywhere then the work done is simply \(F*s\) right

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1since the friction is changing, we need to setup an integral : \[\int_0^1 F\cdot ds\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I don't remember the exact friction formulas tho

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[f_k = \mu_k \times N \]Here, \(f_k = \mu_k \times m g\) where \(m \) is the mass on which friction acts.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[w=\int_0^1 xmg~dx\] ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.1oh right not entire block is above friction surface as it moves

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So\[W_{net} = mg\int_{0}^1 (x  x^2)dx = mg \left[\frac{x^2}{2}  \dfrac{x^3}{3}\right]_{x = 0}^{x=1}\]\[= mg \left[\frac{1}{2}  \frac{1}{3}\right] = \frac{mg}{6}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That isn't correct. :

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1whats the correct answer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we did not factor in the horizontal force F in our calculation

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2It'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
 one year ago
Best ResponseYou've already chosen the best response.2Oh yes, but is it relevant?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Isn't the work done by friction independent of the magnitude of force anyway?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Oh right, we don't need net force on block, just the friction

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So our answer is coming out to be\[\frac{2 \cdot 10}{6} J= \frac{10}{3}J\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1don't trust me on these things, but wait, what is the numerical value of given F

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2No numerical value is given. :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1oh then it shouldnt matter as you said haha

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Tagging @Kainui to check our work.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2The friction at each point is also different though.dw:1433253462744:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2We have done the integral assuming that friction remains the same at A and B and all other such points, right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Again, I'm not sure. The integral may compensate for that, but still, I think that's our catch.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433253743704:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Let'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 \(M(1x)\).\[dW_f = \mu_k N \]\[= x\cdot dm\cdot g\cdot dx\]So net work:\[W_{net, f}=g \int_{0}^{1} \int_0^{M(1x)} xdmdx\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1do you have 20/3 in options

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes  how did you get that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1The trick is to call "x" as the length of block that is on the friction surface

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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: \[dW =\int_s^1 \bar f_k \cdot d \bar x = \int_s^1 xdm dx = \frac{1}{2}(1s^2)dm\] 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 \[\int_0^1 dW = \int_0^1 \frac{1}{2}(1s^2)m ds\] 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
 one year ago
Best ResponseYou've already chosen the best response.2@Kainui You're actually right. You just forgot the factor of \(g\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Let \(x\) = 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
 one year ago
Best ResponseYou've already chosen the best response.2Oh, wow, that sounds easier. Where did our setup go wrong?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we assumed "x" as a point on surface instead of varying length of the block on the friction surface

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1setting up correct integral is somewhat pain if we treat x as a point on the surface

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I see. The difference in our integrals is that \(\mu_k = x\) gives the correct answer in your case.
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