JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Laws of Motion

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A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10s. Coefficient of kinetic friction is (given,

g = 10 ms^{- 2}

(A)0.60
(B)0.75
(C)0.50✓ Correct
(D)0.25

Answer: (C)

Solution

Starting from rest,

s = \frac{1}{2} a t^{2}

gives

50 = \frac{1}{2} a (10)^{2}

, so the acceleration is

a = 1 m s^{- 2}

. Along the table the applied force less kinetic friction produces this acceleration,

F - μ m g = ma

, that is

30 - μ (5) (10) = 5 (1)

. Then

50 μ = 25

, so the coefficient of kinetic friction is

0.50

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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