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JEE Main 2024 (1 Feb, Shift 1) · Friction Basics

Friction Basics question from JEE Main 2024 (1 Feb, Shift 1)

Q6JEE Main 2024 · 1 FEB · SHIFT 1Medium

Friction Basics

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\text{ms}^{-2} is:

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

Figure

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Answer: (d)

  1. Driving force: weight of 6 kg6\text{ kg} block =60 N= 60\text{ N}.
  2. Friction on trolley fk=0.04×20×10=8 Nf_k = 0.04 \times 20 \times 10 = 8\text{ N}.
  3. a=60−820+6=5226a = \dfrac{60 - 8}{20 + 6} = \dfrac{52}{26}
  4. So the answer is (d) 2 m/s22\text{ m/s}^2.
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