Q.86. Mechanics - Inclined Plane.

Question 86.
An 80 kg skier slides down a 25 degree hill with an initial speed of 1.4 m/s. If the coefficient of kinetic friction between the skis and the snow is 0.07 and the hill is 50m long, then
a) how long does it take the skier to reach the bottom of the hill,
b) what is the skier’s velocity at the bottom of the hill, and
c) if the skier comes to a horizontal stretch of snow at the bottom of the hill with the same coefficient of etic friction, then how long will they slide before coming to a halt?

Answer 86.
Weight of the skier
= mg
= 80 * 9.8
= 784 N.

Components of weight along and normal to the hill are
along the hill = 784 sin25° = 331 N and
normal to the hill = 784 cos25° = 711 N

Frictional force
= normal reaction x coefficient of friction
= 711 x 0.07 = 50 N

Force down the hill
= 331 - 50
= 281 N

Acceleration down the hill
= 281/80
= 3.51 m/s^2

a)
s = ut + (1/2)at^2
=> 50 = 1.4t + (0.5)*(3.51)*t^2
=> t^2 + 0.8t - 28.5 = 0
=> t = (1/2) [- 0.8 + √(0.64 + 114)]
= 4.95 s.

b)
Skier's velocity at the bottom of the hill
v = u + at
= 1.4 + (3.51) * (4.95)
= 18.8 m/s

c)
Frictional force on the horizontal surface
= 80 * 9.8 * 0.07 N
= 55 N
deceleration
= 55/80
= 6.9 m/s^2
v = u - at
=> t = u/a
= 18.8/6.9
= 2.7 s.

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