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/s2)

Options:

A) 5 N

B) 10 N

C) 15 N

D) 20 N

Show Answer

Answer:

Correct Answer: B

Solution:

Taking mA=0.5kg;mB=1Kg Force on block A ……(1) Force acting on block B

F=T+μmAg+μ(mA+mB)g ……(2)

From 1 & 2

F=μmAg+μmAg+μmAg+μmBg

F=3μmAg+μmBg=μg(3mA+mB)

=0.4×10×(3×0.5+1)=10N



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