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(B) Given: mass $m = 0.1 \ kg$,change in momentum $\Delta p = 2 \ kg \ m/s$,time $t = 1 \ s$,$g = 10 \ m/s^2$.Since $\Delta p = F_{net} 	imes t$,we h

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The tension in the string is $3 \ N$

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(B) Given: mass $m = 0.1 \ kg$,change in momentum $\Delta p = 2 \ kg \ m/s$,time $t = 1 \ s$,$g = 10 \ m/s^2$.Since $\Delta p = F_{net} 	imes t$,we have $F_{net} = \frac{\Delta p}{t} = \frac{2}{1} = 2 \ N$.For the block,the net force is $T - mg = F_{net}$,where $T$ is the tension in the string. Since the string is massless and the pulley is frictionless,$T = F$.Thus,$F - mg = 2 \ N \Rightarrow F - (0.1 	imes 10) = 2 \Rightarrow F - 1 = 2 \Rightarrow F = 3 \ N$.So,the tension in the string is $T = F = 3 \ N$.Acceleration $a = \frac{F_{net}}{m} = \frac{2}{0.1} = 20 \ m/s^2$.Displacement $S = \frac{1}{2}at^2 = \frac{1}{2} 	imes 20 	imes (1)^2 = 10 \ m$.Work done by tension $W_T = T 	imes S = 3 	imes 10 = 30 \ J$.Work done against gravity $W_g = mg 	imes S = (0.1 	imes 10) 	imes 10 = 10 \ J$.

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