# 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 $$m$$ 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

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

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

Note by Swapnil Das
1 year, 4 months ago

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- 1 year, 4 months ago

Jai Ho :P

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

- 1 year, 4 months ago

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

- 1 year, 4 months ago

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

- 1 year, 4 months ago

Thanks :)

- 1 year, 4 months ago

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

- 1 year, 4 months ago

- 1 year, 4 months ago

- 1 year, 4 months ago

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

- 1 year, 4 months ago