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Standard XII

Physics

Friction General Understanding

When a horizo...

Question

When a horizontal force of 120 N acts on a body of mass 20 kg and moves it through a distance of 10 m, the kinetic energy gained by the body is 400 J. Then, coefficient of kinetic friction between the body and the surface on which it is moving is ( $g = 10 m s^{- 2}$ )

0.2

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0.4

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Solution

The correct option is C $0.4$
Work Done $= 120 \times 10 = 1200 J$
Work Done by friction $= 1200 - 400 = 800 J$
work done by friction $= μ m g s$
$∴ μ \times 20 k g \times 10 m / s^{2} \times 10 m = 800 J$
$∴ μ = 0.4$

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When a horizontal force of 120 N acts on a body of mass 20 kg and moves it through a distance of 10 m, the kinetic energy gained by the body is 400 J. Then, coefficient of kinetic friction between the body and the surface on which it is moving is (g=10 ms−2)

The correct option is C 0.4Work Done=120×10=1200JWork Done by friction=1200−400=800Jwork done by friction=μmgs∴μ×20kg×10m/s2×10m=800J∴μ=0.4

When a horizontal force of 120 N acts on a body of mass 20 kg and moves it through a distance of 10 m, the kinetic energy gained by the body is 400 J. Then, coefficient of kinetic friction between the body and the surface on which it is moving is ( $g = 10 m s^{- 2}$ )

The correct option is C $0.4$
Work Done $= 120 \times 10 = 1200 J$
Work Done by friction $= 1200 - 400 = 800 J$
work done by friction $= μ m g s$
$∴ μ \times 20 k g \times 10 m / s^{2} \times 10 m = 800 J$
$∴ μ = 0.4$