A 79.5 kg snowboarder heads down a 28.0° hill that has a height of 62.8 m. If the hill is assumed to be frictionless and there is horizontal wind with a force of 98 N acting against the snowboarder, find the speed of the snowboarder as they reach the bottom of the hill using work and energy.

Question

So here is what I did
W = Fdcos(theta)
= 98 * 62.8* cos(28)= 5434.01
Then
v = sqrt (2W/m) = 11.7 m/s

I'm not sure what I did is correct... Please help...

Answer 1

Well, I would have used energy
loss of potential energy = gain in kinetic energy + work done by wind

loss of potential energy = m g h
= 79.5 (9.81) (62.8) Joules
= 48,977 Joules

work done by wind
component of horizontal wind opposite direction of motion = 98 cos 28 Newton

distance traveled = 62.8/sin 28 meter
so work done by wind = 98*62.8*cot 28
= 11,574 Joule

so
(1/2) m v^2 = 48,977 - 11,574

Answer 2

Thank you... :-)