Evaluating time taken

I was trying to create a problem yesterday, something like this:

The yellow region represents a smooth surface, while brown represents the rough one.

  • A block of mass mm enters the rough region while initially traversing with a constant velocity vv. We need to find the time taken by the block to come to rest.

The coefficient of friction between the block and the rough surface varies as k=bxk=bx, where xx is the distance traversed by the block on the rough surface (assuming the intersection of the axes looking lines as the origin) and bb is a positive constant.

As an initial approach, I tried to find out the stopping distance of the block, which goes like this.

mvdvdx=kmg\displaystyle mv\frac { dv }{ dx } =-kmg or mvdvdx=bxmgmv\frac { dv }{ dx } =-bxmg as the only force acting on the block is friction.

mvdv=bmgxdxmv0vdv=bmg0x0xdx\displaystyle mvdv=-bmgxdx\\ m\int _{ v }^{ 0 }{ vdv= } -bmg\int _{ 0 }^{ { x }_{ 0 } }{ xdx }

Giving

v22=bgx022x0=vbg\displaystyle \frac { { v }^{ 2 } }{ 2 } =\frac { bg{ { x }_{ 0 } }^{ 2 } }{ 2 } \\ \\ { x }_{ 0 }=\frac { v }{ \sqrt { bg } }

But couldn't proceed on to evaluate time of motion. Please help me, thanks.

PS: Please rectify if there's any mistake in evaluating stopping distance.

Note by Swapnil Das
2 years, 5 months ago

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No its absolutely correct.Infact the time would be t=π/(2bg)t=π/(2√bg).Good to see you make your own prob.(☺☺)

Spandan Senapati - 2 years, 5 months ago

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Thanks :)

Swapnil Das - 2 years, 5 months ago

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Brilliant Member - 2 years, 5 months ago

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Jai Ho :P

The mathematics was complicating for me, I believe, Thanks for the same. Had you encountered such type of question earlier?

Swapnil Das - 2 years, 5 months ago

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not same but similar which i solved by other method ... first time using it :)

Brilliant Member - 2 years, 5 months ago

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@Spandan Senapati

@shubham dhull

Swapnil Das - 2 years, 5 months ago

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Can you tell where you are exactly stuck......BTW here we need to make use of integraldx/(a2x2)=arcsin(x/a)+cintegral dx/√(a^2-x^2)=arcsin(x/a)+c

Spandan Senapati - 2 years, 5 months ago

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clr or you want me to show it completely how to do ?

Brilliant Member - 2 years, 5 months ago

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Show bhai please :P

Swapnil Das - 2 years, 5 months ago

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