Peter & Margeret Homework: KE, Force & Power

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The discussion focuses on a physics homework problem involving kinetic energy (KE), potential energy (PE), and forces related to two skaters, Peter and Margeret. Key calculations include Peter's initial and final KE, the average force required to increase his velocity, and the combined velocity after a collision with Margeret. The change in PE when Peter lifts Margeret is calculated, along with the power exerted during the lift. Finally, the forces exerted between Peter and Margeret during their separation are analyzed, with corrections noted for accuracy in results.
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Homework Statement



a) Peter, with a mass of 80kg, was skating on a frozen lake with a velocity of 1m/s west. What was kis KE?
b) After seeing the beatiful Margeret, Peter increased his velocity to 3m/s west over a distance of 10m.
bi) What was his final KE?
bii) What average force had to applied to the ice to achieve this new velocity of 3m/s west?
c) Peter, in hsi enthusiasm, failed to slow down sufficiently, and while he was moving with a velocity of 1.5m/s west he collided with Margeret who was moving with a velocity of 0.5m/s west. Peter held Margeret (mass 60kg) and the two moved forward together. What was the velocity with which they moved together?
d) Peter now picked up the 60kg Margeret, lifting her through a vertical height of 0.8m. What was her change on PE?
e) When Peter lifted Margeret, he performed this feat in 1.2s. With what power was he working?
f) At some later stage, with the two skaters moving together at 1m/s west, Margeret and Peter push apart and Margeret moves away at 2.2m/s east.
fi) What force did Peter exert on Margeret if it occurred over a period of 0.5s?
fii) What force did Margeret on Peter during the same 0.5s of time?


Homework Equations



a) KE = 0.5 x m x v^2
bi) KE = 0.5 x m x v^2
bii) average force = (final KE - inital KE)/distance
c) m1u1 + m2u2 = m1v1 + m2v2
d) final PE - initial PE = change in PE
e) P = W/t
fi) F = [m(v-u)]/t and m1u1 + m2u2 = m1v1 + m2v2
fii) F = [m(v-u)]/t


The Attempt at a Solution



a) KE = 0.5 x 80 x 1^2 = 40J
bi) KE = 0.5 x 80 x 3^2 = 360J
bii) average force = (360 - 40)/10 = 321J/m
c) {80x1.5 + 60x0.5 = 140v} {v = 1.07m/s west}
d) change in PE = (60x9.81x0.8) - (60x9.81x0) = 470.88J
e) P = (60x9.81x0.8)/1.2 = 392.4 watts
fi) {80x1 + 60x1 = 80v + 60x-2.2}, therefore v = 3.4m/s
F = [80(3.4-1)]/0.5 = 384N
fii) F = [60(-2.2-1)]/0.5 = -348N
 
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I think all are correct
 
Two typos aside, you're completely correct. :)
(For bii you meant 32 N, correct? And for fii you meant -384N, right?)
Just a tip, for fii, you didn't have to go through the trouble of plugging in the numbers. Newton's third law would have given you the answer straight away, do you understand why?

Oh, and on a side note, unless you've agreed beforehand with your teacher that you round your results off to the second decimal place, make sure you keep track of your truncated results.
 
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