A roller coaster reaches the top of the steepest hill with a speed of 7.4 km/h.

Question

A roller coaster reaches the top of the steepest hill with a speed of 7.4 km/h. It then descends the hill, which is at an average angle of 43° and is 45.0 m long. What will its speed be when it reaches the bottom? Assume µ_k = 0.12.

a) _____m/s

Answer 1

Let

v₁ = 7.4 km/h = 2.06 m/s be the speed at the top,

v₂ (to be found) be the speed at the bottom,

s = 45 m be the length of the hill,

α = 43° be the angle of the hill with the horizontal (I assume),

h = s×sin(α) be the height of the hill.

µ_k = 0.12 be the coefficient of friction,

m (unknown) be the mass of the coaster,

g = 9.81 m/s² be gravitational acceleration,

K₁ = m×v₁²/2 be the kinetic energy at the top,

P₁ = m×g×h be the potential energy at the top,

K₂ = m×v₂²/2 be the kinetic energy at the bottom,

P₂ = 0 be the potential energy at the bottom,

G = m×g×cos(α) be the component of the weight (m×g) normal to the surface of the hill,

F = G×µ_k be the force of friction,

W = F×s be the work friction performs while the coaster is sliding down.

For the calculation of the potential energies P₁ and P2 we chose them to be with respect

to the bottom of the hill.

That is what makes P₂ = 0.

We use conservation of energy: The total energy at the top gets converted to the total energy at the bottom plus the work of friction.

By "total energy" we mean kinetic plus potential energy. We should also take into account the rotational energy of the wheels, but since the problem statement says nothing about the wheels, I assume that we are supposed to ignore that.

K₁ + P₁ = K₂ + P₂ + W.

Substituting

m×v₁²/2 + m×g×s×sin(α) = m×v₂²/2 + m×g×cos(α)×µ_k×s

Dividing by m/2:

(Note that this makes the result independent of the mass of the coaster.)

v₁² + 2×g×s×sin(α) = v₂² + 2×g×cos(α)×µ_k×s

Expressing v2:

v₂ = √(v₁² + 2×g×s(sin(α) - cos(α)µ_k)

Substituting actual values

v₂ = √(2.06² + 2×9.81×45×(sin(43°) - cos(43°)×0.12)) = 23 m/s