Question from Linear Kinetics

Question

 A snowboarder takes speed downhill in order to get up a hill with an 8% incline without pushing. The kinetic friction coefficient between snowboard and snow = 0.10 and air resistance can be neglected. (Hint: a 1% slope rises 1 m vertically for every 100 m of lateral travel.)

a) What is the acceleration of the snowboarder parallel to the slope (down the slope is negative)?

b) How far up the hill will the boarder slide without pushing, if they arrive at the bottom at 60 km/h?

Answer 1

Let

f = 0.10 be the coefficient of friction,

θ be the angle between the incline and the horizontal,

m (unknown) be the mass of the sled,

a (to be computed) be the acceleration of the sled,

g = -9.81 m/s² be gravitational acceleration.

(The gravitational acceleration is negative because it is downwards.)

We are given the incline to be 8%, which means

sin(θ) = 0.08

a)

Please draw a diagram.

The sled is subject to three forces:

1) the force of gravity -- mg.

This force can be decomposed into the vector sum of

a force parallel to the incline -- mgsin(θ), and

a force perpendicular to the incline -- mgcos(θ).

2) The normal force of the ground -- mgcos(θ).

3) Downward force of friction -- mgcos(θ)f.

The net force parallel to the ground is the sum

F = mgsin(θ) + mgcos(θ)f

By Newton's second law the sled is subject to acceleration

a = F/m = gsin(θ) + gcos(θ)f

= g(sin(θ) + f√(1 - sin²(θ))

= 9.81×(0.08 + 0.1×√(1 - 0.08²)) = -1.76 m/s²

b)

Let

v = 60 km/h = 60,000m/3600s = 50/3 m/s be the initial velocity

t (unknown) be the time at which the sled's velocity becomes 0,

s (to be computed) be the distance the sled travels by the time t.

By definition of acceleration

(0 - v) = at

So

t = -v/a

By definition of acceleration

s = vt + at²/2 = -v²/a + a(-v/a)²/2 = -v²/a + v²/2a = -v²/2a

= -(50/3)²/(2×(-1.76)) = 78.9 m