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← JEE Main 2022 · 30 Jun · Shift 1
PhysicsSingle correct+4 / -1Question 36 of 90

JEE Main 2022 · 30 Jun · Shift 1, Physics question 36

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=10g=10 ms−2^{-2})

A 2 kg block pressed against a vertical wall by a horizontal 50 N force, with an upward force F applied to it.
Choose one option
Answer and worked solution

Official answer: D

Limiting friction 0.5×50=250.5\times50=25 N acts downward when the block tends to rise. Largest F: 20+25=4520+25=45 N.

(D)

Topics:Friction