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A block of mass 2 kg is resting on a horizontaltable. A force of 10 N is applied for 4 s on theblock in the horizontal direction. If g 10 ms-1and the cofficient of kinetic friction between theblock and the table 0.2, then thethe net force is.(a) 144J (b)96J (c) 72J (d) 36 Jwork done by

Answer»

Answer: a)144 JExplanation: Friction force= 4 Napplied force=10 NNet force=applied - friction=10-4=6 Nacceleration=Force/ mass= 6 N/2 kg=3 m/s^2distance= 24 mwork done by net force= 6 N* 24 m=144 J

Distance travelled =s=ut+1/2(ať^2)here u =initial speed =0s=0+1/2(3*4*4)=3*4*2=24



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