Question 122: Laws of Motion

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 ms
$$-$$2
)

Accepted Answer

Here F
max
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,
F
max
= mg + f
max
= 2 $$	imes$$ 10 + $$\mu$$N
= 20 + 0.5 $$	imes$$ 50
= 20 + 25
= 45 N

Answer

10 N

Answer

20 N

Answer

25 N

Answer

45 N

Laws of Motion

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