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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Here, $$mA=0.5$$ kg;$$mB=1$$ kg Force on block A  $$T=$$μmAg..............(1)  Force acting on block B $$F=T+$$μ$$mAg+$$μ$$mA+mBg$$...........(2)  From (1)&(2) $$F=$$μ$$mAg+$$μ$$mAg+$$μ$$mAg+$$μmBg $$F=3$$μ$$mAg+$$μ$$mBg=$$μ$$g3$$ $$mA+mB$$ $$=0.4$$×$$10$$×$$(3$$×$$0.5+1)=10$$ N

Friction

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Friction

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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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