A93.5 kg snowboarder starts from rest and goes down a 60 degree slope with a 45.7 m height to a rough horizontal surface that is 10.0 m long and a coefficient of kinetic friction of 0.102. the then snowboarder goes up a slope with angle of 30 degrees up coming to a stop and then sits to stay in the same place. be considered frictionless. a. what is the speed of the snowboarder at the bottom of the 60 degree slope? b. what is the speed of the snowboarder at the bottom of the 30 degree slope? c. what is the final height of the snowboarder above the horizontal surface? until both slopes are smooth and can

a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m

Explanation:

This can be answered with either force analysis and kinematics, or work and energy.

a) Using force analysis, we can draw a free body diagram for the snowboarder.  There are two forces: normal force perpendicular to the slope and gravity down.

Sum of the forces parallel to the slope:

∑F = ma

mg sin θ = ma

a = g sin θ

Therefore, the velocity at the bottom is:

v² = v₀² + 2a(x - x₀)

v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)

v = 29.9 m/s

Alternatively, using energy:

PE = KE

mgh = 1/2 mv²

v = √(2gh)

v = √(2×9.8×45.7)

v = 29.9 m/s

b) Drawing a free body diagram, there are three forces on the snowboarder.  Normal force up, gravity down, and friction to the left.

Sum of the forces in the y direction:

∑F = ma

N - mg = 0

N = mg

Sum of the forces in the x direction:

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Therefore, the snowboarder's final speed is:

v² = v₀² + 2a(x - x₀)

v² = (29.9)² + 2(-9.8×.102) (10 - 0)

v = 29.6 m/s

Using energy instead:

KE = KE + W

1/2 mv² = 1/2 mv² + F d

1/2 mv² = 1/2 mv² + mgμ d

1/2 v² = 1/2 v² + gμ d

1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)

v = 29.6 m/s

c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.

∑F = ma

-mg sin θ = ma

a = -g sin θ

Therefore, the final height reached is:

v² = v₀² + 2a(x - x₀)

(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)

h = 44.7 m

Using energy:

KE = PE

1/2 mv² = mgh

h = v² / (2g)

h = (29.6)² / (2×9.8)

h = 44.7 m