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
Answers
Answered by
1
2 a s = v² - u² => v² = - 2 a s
friction force = f = m a = - x m g
a = - x g
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
v² = 2 μ g s
v = √2μgs
friction force = f = m a = - x m g
a = - x g
v = √(2 x g s)
==================================
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
v² = 2 μ g s
v = √2μgs
Answered by
1
Step-by-step explanation:
final velocity=0
initial velocity=v
work is done against friction
final velocity=0
initial velocity=v
work is done against friction
final velocity=0
initial velocity=v
work is done against friction
final velocity=0
initial velocity=v
work is done against friction
final velocity=0
initial velocity=v
work is done against friction
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