what is the total vertical drop of the track?

Question

If a bobsled with driver (combined mass =300kg) starts at the top of the track with an initial velocity of 2m/s and at the end of the run, the sled is speeding along at 30m/s. Assume the track’s friction consumes 10 percent of the total gravitational energy, what is the total vertical drop of the track?

Answer 1

Initial kinetic energy
Ki = 1/2 . m . v^2 = 1/2 . 300 kg . (2m/s)^2
Ki = 600 J

Final kinetic energy
Kf = 1/2 . 300 kg . (30m/s)^2 = 135 000 J

Difference between final and initial is equal to loss of potential energy (mgh) minus friction energy (Ef)

Kf - Ki = m.g.h - Ef

and Ef = 10% of potential energy so

Kf - Ki = m.g.h - 10%(m.g.h) = 90% m.g.h

h = 100/90 . (Kf - Ki)/(m.g)
h = 100/90 . (135 000 - 600) / (600 . 9.8)
h = 50.79 m

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Note: the previous answer doesn't take the initial kinetic energy of 600J into account and obtains a slightly different but erroneous answer (slightly because 600 is small compared to 135 000).

Answer 2

Using conservation of energy:

m*g*h+.5*m*vi^2=.5*m*v^2-friction work

from the problem statement
m*g*h*.9+.5*m*4=.5*m*30^2
h=(.5*30*30-.5*4)/(.9*9.81)
h=51 m

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