Laws Of Motion Question 387

Question: The masses of the blocks A and B are 0.5 kg and 1 kg respectively. These are arranged as shown in the figure and are connected by a massless string. The coefficient of friction between all contact surfaces is 0.4. The force, necessary to move the block B with constant velocity, will be $ (g=10m/s^{2}) $

Options:

A) 5 N

B) 10 N

C) 15 N

D) 20 N

Show Answer

Answer:

Correct Answer: B

Solution:

Taking $ m _{A}=0.5kg;m _{B}=1Kg $ Force on block A ……(1) Force acting on block B

$ F=T+\mu m _{A}g+\mu (m _{A}+m _{B})g $ ……(2)

From 1 & 2

$ F=\mu m _{A}g+\mu m _{A}g+\mu m _{A}g+\mu m _{B}g $

$ F=3\mu m _{A}g+\mu m _{B}g=\mu g(3m _{A}+m _{B}) $

$ =0.4\times 10\times (3\times 0.5+1)=10N $



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