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Question

A small particle slides along a track with elevated ends and a flat central part as shown in figure. The flat part has a length of 3m and the curved portions of the track are frictionless. For the flat part, coefficient of kinetic friction is μ = 0.2. The particle is released at a point A, which is at a height 0.3 m above the flat part of track. The position where the particle finally come to rest is :
294017_24ebe8c23c9d49fcbae14700df6f1e52.png

A
left to mid point of the flat part
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B
right to the mid point of the flat part
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C
mid point of the flat part
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D
none of these
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Solution

The correct option is C mid point of the flat part
Velocity of block when it enters into the flat surface -
12mu2=mgh=m(10)(0.3) u2=6
Now from flat part, its deacceleration due to friction is a=μg=0.2(10)=2m/s2

Using the equation - v2=u2+2as
0=62(2)s s=1.5m

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