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Question

A horizontal net force of 29.8 N is applied to a 10.8 kg mass initially at rest on a very long frictionless surface. If this net force acts on this object for 2.20 s, calculate the force required to stop this mass in a distance of 1.00 m

Answer 1

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final velocity V  = a*t = F/m*t = 29.8/10.8*2.2 = 6.070 m/sec 

KE + F*d = 0

F = -KE/d = -10.8/2*6.070^2/1 = -199 N 

Answer 2

Notice that N*meter = kg*m²/s² = Energy in Joules

F•d = F'•1 where F' is the unknown force.

29.8N/10.8kg = a = 149/54 m/s²

d = ½at² = (149/108)*2.2s² = 6.68m is the distance the 10.8kg was pushed. Vf = Vo + a*t = 0 + 149/54 *2.2 = 6.07m/sKE = ½*m*V² = ½*10.8*6.07² = 199J = F' •1mF' = 199N <<<<Try it another way. The 29.8N force x the distance 6.68m = Force F' * 1m29.8N*6.68m = 199J = 199N•1m Checks