Question 51: Laws of Motion

Question

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
]

Accepted Answer

$$a = {F \over {M + m}}$$
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
$$ 	herefore $$ F
max
= $${3 \over 7}\left( {0.5 + 4.5} ight) 	imes 9.8$$ = 21 N

Laws of Motion

Integer / Numeric Answer Type