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

- schrodinger

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- ParthKohli

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- ParthKohli

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- ParthKohli

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## More answers

- 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

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

- ParthKohli

Yes. Can it be done without double-integration?

- ParthKohli

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

- ganeshie8

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

- 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

Just looking for your confirmation.

- ganeshie8

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

- 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

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

- ganeshie8

I don't remember the exact friction formulas tho

- 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

\[w=\int_0^1 xmg~dx\]
?

- 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

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

- 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

That isn't correct. :|

- ganeshie8

whats the correct answer

- ganeshie8

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

- 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

Oh yes, but is it relevant?

- ParthKohli

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

- ganeshie8

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

- ParthKohli

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

- ganeshie8

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

- ParthKohli

No numerical value is given. :P

- ganeshie8

oh then it shouldnt matter as you said haha

- ParthKohli

Tagging @Kainui to check our work.

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

|dw:1433253743704:dw|

- 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

do you have 20/3 in options

- ParthKohli

Yes - how did you get that?

- ganeshie8

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

- 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

@Kainui You're actually right. You just forgot the factor of \(g\).

- 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

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

- ganeshie8

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

- ParthKohli

Oh...

- ganeshie8

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

- ParthKohli

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

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