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Q.

Two blocks A and B of mass 10 kg and 20 kg respectively are placed as shown in figure. Coefficient of friction between all the surfaces is 0.2 (g = 10m /s2 ) . Then

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a

acceleration of block B is 2.6 m/ s2

b

tension in the string is 132 N

c

acceleration of block B is 4.7 m /s2

d

tension in the string is 306 N

answer is A, D.

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Detailed Solution

free body diagram of A and B is

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For A
T cos300 =N
T sin 300 = 0.2N +100
Solving these two equations we get
 306 Newton and N 265 Newton
For B
a=mBg0.2N0.2NmB=2000.4(265)20=4.7m/s2

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