A worker is pushing a heavy box across a concrete floor. At first, he has to pus
ID: 1328916 • Letter: A
Question
A worker is pushing a heavy box across a concrete floor. At first, he has to push as hard as he can to get it moving. Once it is moving with a reasonable speed, he still has to push on it but not as hard. Finally, to get it to stop, he simply stops pushing and lets it stop by itself. We'll call the three different parts of his time with this box as "phases". In phase 1 he's pushing hard to get it moving; in phase 2 he's pushing moderately hard to keep it moving; and in phase 3 he's not pushing at all. Assume he always pushes horizontally.
Assume that in each phase he pushes with a constant force. Phase 1 lasts 5 seconds, phase 2 lasts 10 seconds, and phase 3 lasts 2 seconds. The box has a mass of 20 kg and in phase 2 he can keep it going steadily by exerting a force of 80N. How much force did he exert in phase 1?
Explanation / Answer
from the given data, kinetic friction force = 80 N
mue_k*m*g = 80
mue_k = 80/(m*g)
= 80/(20*9.8)
= 0.408
let mue_s is the static friction.
Fnet = F_applied - kinetic friction
m*a = mue_s*m*g - mue_k*m*g
a = mue_s*g - mue_k*g ------------(1)
let v is the speed gained in phase 1
box is pushing with constant speed in phase 2.
in phase it comes to stop.
in phase 3, accelration, a = -g*mue_k
= -9.8*0.408
= -4 m/s^2
vf = v + a*t
0 = v -4*2
==> v = 8 m/s
acceleration of box in the first phase, a = (8 - 0)/5
= 1.6 m/s^2
now from equation 1
1.5 = mue_s*9.8 - 0.408*9.8
mue_s = (1.5 + 0.408*9.8)/9.8
= 0.56
so, Applied force in phase 1, F_applied = mue_s*m*g
= 0.56*20*9.8
= 109.76 N <<<<<<<<<<<<<<<<<<<<<<<<<---------------Answer
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