Asked by Khalid
A ski jumper glides down a 30.0 degrees slope for 80.0 ft before taking off from a negligibly short horizontal ramp. If the jumpers takeoff speed is 45.0 ft/s, what is the coefficient of kinetic friction between skis and slope?
Answers
Answered by
Damon
friction force up slope = mu m g cos 30
work done by friction = -80 mu m g cos 30
drop in height = 80 sin 30 = 40 ft
potential energy drop = m g (40 ft)
ke = (1/2) m v^2
so
(1/2)m(45)^2 = 40 m g - 80 mu m g cos 30
1012 = 40(32) - mu * 32 * 69.3
1012 = 1280 - 2218 mu
mu = (1288 - 1012)/2218 = .124
work done by friction = -80 mu m g cos 30
drop in height = 80 sin 30 = 40 ft
potential energy drop = m g (40 ft)
ke = (1/2) m v^2
so
(1/2)m(45)^2 = 40 m g - 80 mu m g cos 30
1012 = 40(32) - mu * 32 * 69.3
1012 = 1280 - 2218 mu
mu = (1288 - 1012)/2218 = .124
Answered by
Anonymous
Ah
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.