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:
5 N
B) 10 N
C) 15 N
D) 20 N
Show Answer
Answer:
Correct Answer: B
Solution:
Taking $ m _{A}=0.5,kg;m _{B}=1,kg $ Force on block A … (1) Force acting on block B
$ F=T+\mu m _{A}g+\mu m _{B}g $ … (2)
From 1 and 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 $
BETA
