A body of mass m moves with the velocity v on a surface whose friction co-efficient is x . If the body covers distance s before coming to rest, then v will be (A) square root of 2xgs (B)square root of xgs (C) square root of xgs/2 (D) square root of 3xgs explain with answer

1
by iitjeepraveen

2014-12-21T22:34:19+05:30

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from the laws of motion:       2 a s =  v² - u²

=>       v² =  - 2 a s    as  initial velocity = v  and  final is 0.

friction force = f =  m a = - x m g

=>     a = - x g  substituting in the above equation for v, we get

v = √(2 x g s)

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Kinetic energy of the body = 1/2 * m * v²

Work done by friction when it stops the body = Force * distance

= (μ m g) * (- s)

work done by friction = - initial kinetic energy + final kinetic energy

- μ m g s =  0 - 1/2 m v²

hence,        v² = 2 μ g s

so         v = √(2μgs)