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 \(m\) enters the rough region while initially traversing with a constant velocity \(v\). 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 \(k=bx\), where \(x\) is the distance traversed by the block on the rough surface (assuming the intersection of the axes looking lines as the origin) and \(b\) is a positive constant.

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

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

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

Giving

\(\displaystyle \frac { { v }^{ 2 } }{ 2 } =\frac { bg{ { x }_{ 0 } }^{ 2 } }{ 2 } \\ \\ { x }_{ 0 }=\frac { v }{ \sqrt { bg } }\)

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.

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The mathematics was complicating for me, I believe, Thanks for the same. Had you encountered such type of question earlier? – Swapnil Das · 5 months, 2 weeks ago

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– Brilliant Member · 5 months, 2 weeks ago

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

No its absolutely correct.Infact the time would be \(t=π/(2√bg)\).Good to see you make your own prob.(☺☺) – Spandan Senapati · 5 months, 2 weeks ago

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– Swapnil Das · 5 months, 2 weeks ago

Thanks :)Log in to reply

clr or you want me to show it completely how to do ? – Brilliant Member · 5 months, 2 weeks ago

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– Swapnil Das · 5 months, 2 weeks ago

Show bhai please :PLog in to reply

@Spandan Senapati

@shubham dhull – Swapnil Das · 5 months, 2 weeks ago

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– Spandan Senapati · 5 months, 2 weeks ago

Can you tell where you are exactly stuck......BTW here we need to make use of \(integral dx/√(a^2-x^2)=arcsin(x/a)+c\)Log in to reply