JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 27)
Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is $${3 \over 7}$$. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is ___________ N. (Round off to the Nearest Integer) [Take g as 9.8 ms$$-$$2]
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Answer
21
Explanation
_17th_March_Morning_Shift_en_27_2.png)
$$a = {F \over {M + m}}$$
f = m$$a$$ = m$${F \over {M + m}}$$
m$${F \over {M + m}}$$ $$ \le $$ $$\mu $$mg (for no slipping)
$$ \Rightarrow $$ F $$ \le $$ $$\mu $$(m + M)g
$$ \therefore $$ Fmax = $${3 \over 7}\left( {0.5 + 4.5} \right) \times 9.8$$ = 21 N
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