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Mass of block A is 100 kg  and that of block B is  200 kg. As shown in figure, the block A is attached to a string tied to the wall. The coefficient of static friction between blocks A and B is (μs)A=0.2 and the coefficient of static friction between block B and floor is (μs)B=0.3. Then calculate the minimum value of force F required to move the block B. (g=10 m/s2)



A
980 N
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B
1000 N
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C
1100 N
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D
1200 N
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Solution

The correct option is C 1100 N

From the FBD of A:
T=f1........(1)
From FBD of B:
F=f1+ f2......(2)

Since we want the minimum value of F (for just slipping condition), we will take the limiting values of friction.
f1=(μs)A mA g .........(3)
and f2=(μs)B(mA+mB)g ......(4)

Substituting (3) and (4) in (2),
F=f1+f2
=(0.2)(100g)+(0.3)(300g)=110 g=1100 N

Physics

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