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A Body Slipping on a Rough Horizontal Plane Moves with a Deceleration of 4.0 M/S2. What is the Coefficient of Kinetic Friction Between the Block and the Plane? - Physics

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प्रश्न

A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s2. What is the coefficient of kinetic friction between the block and the plane?

योग
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उत्तर

Let m be the mass of the body.



From the free body diagram,
R − mg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
⇒ R = mg                 (1)
Again ma − μkR = 0
(where μk is the coefficient of kinetic friction and a is deceleration)
or ma = μkR
From Equation (1),
ma =  μkmg
⇒ a = μkg
⇒ 4 = μkg
`=>muk=4/g=4/10=0.4`

Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.

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अध्याय 6: Friction - Exercise [पृष्ठ ९७]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 6 Friction
Exercise | Q 1 | पृष्ठ ९७
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