Question

A 54.5-kg skateboarder starts out with a speed of 1.95 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 5.95 m/s.

a) Calculate the change (ΔPE = PEf - PE0) in the gravitational potential energy.
b)How much has the vertical height of the skater changed?
c)Is the skater above or below the starting point?

Answer 1

initial Ke = (1/2)(54.5)(1.95)^2
= 104 J

energy added = 80-265 = - 185 J
so
if no change in elevation final ke would be
104-185 = - 81 J

actual final ke = (1/2)(54.5)(5.95)^2
= 965 J

so the gravitational PE loss (dropped down to gain speed)
= 965 + 81 = 1046 J
or
delta PE = -1046 J
if height drop = h
m g h = 1046
h = 1046 / (9.81*54.5) = 1.96 meters drop

the height went down below the start of course

Answer 2

^^^^^^^^^^^^WRONG^^^^^^^^^^^^

Answer 3

First find the initial change in kinetic energy = 1/2 * m * v(initial)^2
1/2 * 54.5 * 1.95^ 2 = 103.6J
Find final kinetic energy = 1/2 * m * v(final)^2
1/2 * 54.5 * 5.95^2 = 964.7J

We know that with non-conservative energy the equation:
Net External Work = change in kinetic energy - (-change in potential energy)

the external work is 80-265 = -185J

so, using the equation above, DPE = change in potential Energy
-185 = (964.7 - 103.6) - (-DPE)
a.) DPE = -1046.1

DPE = PE(final) - PE(initial)
Because the state boarder starts at height 0 (on the ground), PE(initial) is 0

PE(final) = mgh(final) = DPE
-1046.1 = 54.5 * 9.8 * h
b.) h = -1.96m, (this is the final height)

c.) because the final height is less than the initial, the skater is below where he started with respect to height