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Answer to Question #3611 in Mechanics | Relativity for Vikash

Question #3611
A force is applied on a body ,mass 50 kg, for 8 sec. After that the body cover 80m distance in 5 sec in coming to rest.Find the force applied.
Expert's answer
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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