A rectangular box of length 3 2 m is initially travelling at 2 m/s with its entire length over the smooth (i.e. perfectly frictionless) blue surface (shown above). The box gradually moves onto a rough surface (grey) and stops at some point before it has entirely moved on to the rough surface. The coefficient of sliding friction between the block and the rough surface is 0 . 7 .
What length of the block (in meters) remains over the smooth surface when the block stops moving?
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The ODE method is not needed for this problem. However if the question had asked how long will the block take to stop? then this will method can yield the right result.
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I'm just using the simple harmonic motion's concept.
The method is perfect
Let x be the distance travelled by the block over the rough surface. The frictional force acting will therefore be 2 3 x μ g m
Let L be the distance travelled over the rough surface. The block will come to rest when the kinetic energy balances the frictional work. The work done is the integral of the frictional force from 0 to L.
This is 4 3 L 2 μ g m which must equal the kinetic energy of 2 m v 2
This gives L 2 as 2 1 8 and thus the answer as 3 2 − L which is 0 . 0 4 9 5
Exactly sir!
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My intention is that this version won't have the unfortunate feature of the right answer from the wrong analysis.
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Hmm.. Nice intention. Its correct. Why should anyone get it correct?
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@Md Zuhair – In the previous version of this problem by Steven Chase it turned out that by chance the you could get the right answer by pretending all the friction happens at a point directly under the centre of mass instead of linearly increasing with the distance of the block on the rough surface.
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@Ed Sirett – Ya. But that time also, When Steven sir uploaded it, It had 3 parts, One of them was my idea! Well, I did it with proper method of integration in all of them and this one too!
Great question and beautiful method. (+1!)
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