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A small block slides along a track with ...

A small block slides along a track with elevated ends and a flat central part as shown in figure. The flat portion BC has a length `l=3.0 m`, The curved portions of the track are frictionless. For the flat part, the coefficient of kinetic friction is `mu_(k) = 0.20`, the particle is releast at point A which is at height `h = 1.5 m` above the flat part of the track. Where does the block finally comes to rest?

Text Solution

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As initial Mechanical Energy (ME) of the particle is mgh and final is zero, so loss in ME = mgh. This ME is lost in doing work against friction in the flat part, so loss in ME = WD against friction
`mgh = mu mgs`

i.e., `s=(h)/(mu)=(1.5)/(0.2)=7.5m`
After starting from B the particle will reach C and then will rise up till the remaining KE at C is converted into potential energy. It will then again descend and at C will have the same value as it had when ascending, but now it will move from C to B , the same will be repeated and finally the particle will come to rest at E such that `BC+CB+BE=7.5=3+3+BE=7.5`
B = 1.5
So the particle comes to rest at the centre of the flat part.
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