ride4life
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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