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Q.

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 ( in N ) necessary to move the block B with constant velocity will be (g=10 m/s2)

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Detailed Solution

Here, mA=0.5 kg;mB=1 kg Force on block A T=μmAg..............(1) Force acting on block B F=T+μmAg+μmA+mBg...........(2) From (1)&(2) F=μmAg+μmAg+μmAg+μmBg F=3μmAg+μmBg=μg3 mA+mB =0.4×10×(3×0.5+1)=10 N

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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 ( in N ) necessary to move the block B with constant velocity will be (g=10 m/s2)