SATHEE: Laws Of Motion Question 336

Laws Of Motion Question 336

Question: The two blocks, $m = 10 k g$ and $M = 50 k g$ are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to

Options:

A) 100 N

B) 50 N

C) 240 N

D) 180 N

Show Answer

Answer:

Correct Answer: C

Solution:

[c]As m would slip in vertically downward direction, then $m g = μ N$

$\Rightarrow N = \frac{m g}{μ} = \frac{100}{0.5} = 200 N e w t o n$

Same normal force would accelerated M, thus $a_{M} = \frac{200}{50} = 4 m / s^{2}$ Taking $m + M$ as system $F = (m + M) 4 = 240 N$

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Laws Of Motion Question 336

Question: The two blocks, m=10kg and M=50kg are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to

Options:

Answer:

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Question: The two blocks, $m = 10 k g$ and $M = 50 k g$ are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to