JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Laws of Motion

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Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04 , the acceleration of the system in

ms^{- 2}

is :
(Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

(A)3
(B)4
(C)2✓ Correct
(D)1.2

Answer: (C)

Solution

For the block (20 kg) on the trolley connected over a pulley to the hanging 6 kg block, with an applied 60 N force and kinetic friction under the trolley:

f_{k} = μ N = 0.04 \times 20 g = 0.04 \times 20 \times 10 = 8 N

Applying Newton's second law to the whole system (net driving force minus friction, total mass 26 kg):

a = \frac{60 - 8}{20 + 6} = \frac{52}{26} = 2 m/s^{2}

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option

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