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Question

The initial velocity of a body moving on a rough horizontal surface having coefficient of kinetic friction 0.5 is 15 m/s. Distance after which it will come to rest is s1. If coefficient of kinetic friction is reduced to 0.3, distance travelled by the body before coming to rest is s2. Find s1+s2. (Take g=10 m/s2)

A
52.5 m
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B
57 m
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C
60 m
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D
None of the above
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Solution

The correct option is C 60 m
Due to friction, the body undergoes retardation of a=μg
where μ = coefficient of kinetic friction
Also u=15 m/s and v=0 (as it comes to rest)

From kinematics,
v2=u22as
0=u22as
s=u22μg
Therefore,
s1=1522×0.5×10=22.5 m
s2=1522×0.3×10=37.5 m

s1+s2=60 m

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