A 50 kg body is applied with a force for 8 sec. After that the force is removed but the body continous in the motion for 5 sec and cover a displacement of 50m. Find the force.
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Expert's answer
2011-07-18T07:52:22-0400
Index “1” means first part of path and index “2” the second part V is the velocity at time 8sec Ff is a friction force
F1 = F-Ff = ma1 a1 = (F-Ff)/m V = a1 t2 F2 = Ff = ma2 a2 = F2/m V = a2 t2 = F2/m t2 (*) a1 t1 = a2 t2 (F-Ff)/m t1 = Ff/m t2 F = Ff (1+t2/t1 ) (**) A2 = Ff S2 E1 = (mV2)/2 A2 = E1 Let put here the V from (*) Ff S2 = m 1/2 Ff2/m2 t22 so, Ff = (2mS2)/(t22 ) (**): F = (2mS2)/(t22 ) (1+t1/t2 ) = (2*50*80)/52 (1+8/5) = 832 N
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