Work done by friction on smaller block

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

If the two blocks were to move together when the external force of magnitude mg is applied , the acceleration would be mg/3m = g/3 . This acceleration would have to be provided by friction . Value of friction should be mg/3 . But maximum value of friction is mg/4 .

This means there will be slipping .Upper block will move but its displacement will be less than the lower block .

First three options are ruled out . I think option 4) is right.

Is that correct ?

 

[Charles Link]

Yes, you have it correct. There will be slipping. So what is the net force on the lower block? Next, what is the acceleration of lower block? How long does it take the lower block to travel the distance $d$? Then, go to the upper block. You know the horizontal force on it from friction. From that you can compute the acceleration of the upper block. The distance traveled can also be computed because the time was already computed. Computing the work done on it is then quite simple. I believe the answer is indeed none of the above, but what answer do you get for the work done on the upper block? See if you get the same answer I did, but as a Homework Helper, I'm not allowed (by Physics Forums rules) to give you the answer first. :) $\\$ Incidentally, (3) looks like a possibility, because the (horizontal) force on the lower block involves the applied force minus the frictional force from the upper block. When computed over the distance $d$, (3) could certainly be a possibility, because the first block has that much less energy because of the frictional force from the second block. The above prescription gives a different answer though, if my arithmetic is correct, because the frictional forces in this problem will result in some heating, rather than going completely into kinetic energy of motion. Thereby block 2 acquires less energy than answer (3) if my arithmetic is correct. I'm curious to see if you get the same result. $\\$ To say it another way, the horizontal force of $mg$ that is applied to the lower block over a distance $d$ involves a certain amount of work input to the system. Does all of this work get converted to kinetic energy, or is there some energy loss?

 

[Jahnavi]

but what answer do you get for the work done on the upper block?

Time t taken by the lower block for displacement d = 4√(d/3g)

Displacement of the upper block in time t = 2d/3

Work done by friction on upper block = mgd/6

 

[Charles Link]

Very good. :) That agrees with the answer that I got.

 

[Jahnavi]

But kinetic energy gained by the upper block will be less than mgd/6 because of loss in thermal energy ?

If there were no slipping ( two blocks move together ) then work done by friction would be equal to kinetic energy gained by upper block ?

 

[Charles Link]

But kinetic energy gained by the upper block will be less than mgd/6 because of loss in thermal energy ?

If there were no slipping ( two blocks move together ) then work done by friction would be equal to kinetic energy gained by upper block ?

You computed it correctly. The final kinetic energy of the upper block is $mgd/6$. The final kinetic energy of the lower block is $(3/4) mgd$. The work that was input to the system by the applied horizontal force is $mgd$. So how much heat was generated during the process? $\\$ If you just looked at the energy input to the system, and the energy acquired by the lower block, you might make an uneducated guess that (3) $mgd/4$ could be the answer to this problem, but (3) is, as we see, incorrect, and the correct answer is $mgd/6$.

 

[haruspex]

But kinetic energy gained by the upper block will be less than mgd/6 because of loss in thermal energy ?

No, you already allowed for that. Try calculating the work done by the larger block.

Note that the question is worded poorly. It asks for the work done on the smaller block "by friction ... by the larger block". It cannot be both. There is the work done on the smaller block by friction, and the work done against friction by the larger block. The lost work is the difference of the two.

 

[Jahnavi]

There is the work done on the smaller block by friction, and the work done against friction by the larger block. The lost work is the difference of the two.

Would it be correct to say that energy dissipated (thermal energy generated ) is equal to work done against friction in the total system ( both the blocks ) ?

 

[haruspex]

Would it be correct to say that energy dissipated (thermal energy dissipated ) is equal to work done against friction in the total system ( both the blocks ) ?

Yes... mgd/4 done against friction by the larger and -mgd/6 against friction by the smaller.

 

[Charles Link]

Would it be correct to say that energy dissipated (thermal energy generated ) is equal to work done against friction in the total system ( both the blocks ) ?

So try to answer the question: How much heat is generated in the above process as the lower block has moved a distance $d$?

 

[Jahnavi]

So try to answer the question: How much heat is generated in the above process after the lower block has moved a distance $d$?

mgd/4

 

[Charles Link]

mgd/4

Incorrect. The upper block received some kinetic energy. The missing energy gets converted to heat. How much energy is missing? See post 6.

 

[Jahnavi]

Incorrect. The upper block received some kinetic energy. The missing energy gets converted to heat. How much energy is missing? See post 6.

I understand your point .

I am thinking differently . Please see my post #8 .

Work done by friction on lower block is -mgd/4 . Isn't this the net heat energy dissipated ?

 

[haruspex]

I understand your point .

I am thinking differently . Please see my post #8 .

Work done by friction on lower block is -mgd/4 . Isn't this the net heat energy dissipated ?

The lower block does work mgd/4 against friction, and that is all the work that the lower block does. The friction does work mgd/6 on the smaller block. So what is the net work done against friction?

 

[Jahnavi]

The lower block does work mgd/4 against friction, and that is all the work that the lower block does. The friction does work mgd/6 on the smaller block. So what is the net work done against friction?

mgd/12

 

[Charles Link]

I understand your point .

I am thinking differently . Please see my post #8 .

Work done by friction on lower block is -mgd/4 . Isn't this the net heat energy dissipated ?

Post 8, if I am reading it properly, is incorrect. An applied horizontal force puts an input of work to the system. In the optimal case, all of that work would be converted to kinetic energy. $\\$ Meanwhile, the lower block applies a horizontal force to the upper block that is equal and opposite the horizontal force from the upper block on the lower block. That force is $mg/4$. The force on the lower block persists over the distance $d$, resulting in a decrease in energy of the lower block by the amount $W=mgd/4$. The upper block does not, however, acquire this amount of kinetic energy, because its applied force on the upper block only goes a distance of $(2/3)d$.

 

[Jahnavi]

@haruspex , @Charles Link please confirm one thing .

Is heat energy dissipated in the system ( both the blocks ) equal to mgd/12 ?

 

[Charles Link]

That is correct. That would result in a rise in temperature of the blocks.

 

[Jahnavi]

@haruspex , @Charles Link

I would like to understand this bit mathematically . Let us take one block at a time .

Please consider only the upper block .

Work done by friction on upper block = mgd/6

Final kinetic energy of upper block = mgd/6

Agree ?

 

[Charles Link]

@haruspex , @Charles Link

I would like to understand this bit mathematically . Let us take one block at a time .

Please consider only the upper block .

Work done by friction on upper block = mgd/6

Final kinetic energy of upper block = mgd/6

Agree ?

Yes, that is correct.

 

[Jahnavi]

OK .

Now consider the lower block .

Work done by external force = mgd

Work done by friction on lower block = -mgd/6

Final kinetic energy of lower block = 3mgd/4

Agree ?

 

[Charles Link]

OK .

Now consider the lower block .

Work done by external force = mgd

Work done by friction on lower block = -mgd/6

Final kinetic energy of lower block = 3mgd/4

Agree ?

Incorrect. The Frictional force is $F_f=.25 mg$ and the lower block travels a distance $d$. This means the work done by friction on the lower block is $-mgd/4$. $\\$ Suggestion: You might want to work a similar problem, but one where the force and acceleration was such that there was no sliding of the upper block. In that case the work put into the system by the applied force would get completely converted to kinetic energy. $\\$ In the case at hand, we can compute the distance of sliding, and it is $d_s=d-(2/3)d=(1/3)d$. That multiplied by the frictional force of $F_f= mg/4$ gives, in fact, the lost energy of $E=(1/12)mgd$.

 

[Jahnavi]

OK .

You must be bored of explaining things to me

Here is a question for you .

Question -Choose the correct option in the OP if the coefficient of friction is 0.5 instead of 0.25 ?

If your answer is option d) , please tell the answer .I might not help you if you don't show an attempt as that will be against the forum rules

 

[Charles Link]

OK .

You must be bored of explaining things to me

Here is a question for you .

Question -Choose the correct option in the OP if the coefficient of friction is 0.5 instead of 0.25 ?

If your answer is option d) , please tell the answer .I might not help you if you don't show an attempt as that will be against the forum rules

I get $mgd/3$. And the bottom one has $mgd-(1/3)mgd=(2/3)mgd$. Energy is conserved with no heating. The only sliding was the frictionless sliding of the bottom block on the table.

 

[Jahnavi]

I get $mgd/3$.

Wrong answer ! Please try again

 

[Charles Link]

Wrong answer ! Please try again

With a coefficient of friction of $.5$, it doesn't mean the frictional force is necessarily $F_f=.5 \, mg$. That is the maximum friction you could get. For the case where you have sufficient friction, you compute the acceleration of the system $a=F/m_{total}=mg/(3m)=g/3$. The force on the upper block is $mg/3$, and we have as much as $mg/2$ available for frictional force if necessary. $\\$ I don't get them right every time, but this one I think I got correct. :) It travels a distance $d$ so the answer is $mgd/3$. $\\$ The way I first did it, I knew energy must be conserved with no sliding of friction present. Input was $mgd$, and it gets split with two parts to the lower block, and one part to the upper.

 

[Jahnavi]

Sorry ! Calculation mistake on my part

You are right .

 

[Jahnavi]

Thanks a lot Charles ! It was a nice interaction .