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JEE Main 2022 (30 Jun, Shift 1) · Friction Basics

Friction Basics question from JEE Main 2022 (30 Jun, Shift 1)

Q13JEE Main 2022 · 30 JUN · SHIFT 1Medium

Friction Basics

A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be:

(Given: g = 10 ms−{}^{-}2)

Figure

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Answer: (d)

  1. Normal force N=50N = 50 N, so limiting friction =0.5×50=25= 0.5 \times 50 = 25 N.
  2. When the block is about to move up, friction acts downward along with the weight.
  3. Fmax=mg+μN=20+25F_{max} = mg + \mu N = 20 + 25
  4. Fmax=45F_{max} = 45 N, so the answer is (d).
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