Question

A particle slides along a track with elevated ends and a flat central part as shown in Fig.
The flat part has a length l=3.0 m. The curved portions of the track are fricionless. For the
flat part the coefficient of kinetic friction is uk = 0.20, the particle is released at point A which is at height h = 1.5 m above the flat part of the track. Where does the particle finally come to rest?​

Answer 1

AnSwer :

As initial MEvof the particle is mg and final is 0 , 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

${ \sf{mgh = \mu \: mgs}}$

$\sf{.i.e., \: \: s = \dfrac{h}{ \mu} = \dfrac{1.5}{0.2} = 7.5m }$

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 hd 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

BE = 7.5 - 6

BE = 1.5

So, the particle comes to rest at the centre of the flat part

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