JEE Main 2022 — Laws Of Motion Question with Solution
From: JEE Main 2022 (Online) 30th June Morning Shift
Question
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 ms2)
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: D
Step-by-step explanation
Here Fmax force is trying to move the block upward so friction force will be applied towards downward.
Along horizontal direction,
N = 50 N
Along vertical diretion,
Fmax = mg + fmax
= 2 10 + N
= 20 + 0.5 50
= 20 + 25
= 45 N
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