JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 25)
The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be _________ N. [g = 10 ms$$-$$2]
Answer
25
Explanation
Given, coefficient of static friction, $$\mu$$s = 0.2
_24th_February_Morning_Shift_en_25_1.png)
Various forces acting on block are shown below
_24th_February_Morning_Shift_en_25_2.png)
Frictional force $$\le$$ mg
$$\Rightarrow$$ N $$\times$$ 0.2 $$\le$$ 5
$$\Rightarrow$$ N $$\le$$ 25
$$\therefore$$ Magnitude of horizontal force, F = N = 25 N
Various forces acting on block are shown below
Frictional force $$\le$$ mg
$$\Rightarrow$$ N $$\times$$ 0.2 $$\le$$ 5
$$\Rightarrow$$ N $$\le$$ 25
$$\therefore$$ Magnitude of horizontal force, F = N = 25 N
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