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a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at the bottom is 15 m/s. what is the lenght of the incline? how long does it take
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Sep 1, 2017
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a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at the bottom is 15 m/s. what is the lenght of the incline? how long does it take him to reach the bottom?
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... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m
(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry
... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
JayD
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Sep 5, 2017
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JayD
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