An extreme skier, starting from rest, coasts down a mountain slope that makes an angle 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 13.8 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 4.50 m below the edge. How fast is she going just before she lands?

Question

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 13.8 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 4.50 m below the edge. How fast is she going just before she lands?

Answer 1

First calculate the potential energy decrease per unit mass,

delta PE = g *(13.8 sin25 + 4.50)
Subtract from that the work done against friction while on snow (per unit mass),
g*cos25*0.20*13.8

The result should be the kinetic energy per unit mass when she lands, which is V^2/2

Use that to calculate the final speed, V

Answer 2

To calculate the speed at which the skier lands, we need to find her speed at the edge of the cliff and then use that to determine her speed when she lands 4.50 m below.

First, let's find the skier's speed at the edge of the cliff. We can use the principle of conservation of mechanical energy to solve this problem.

The initial mechanical energy of the skier is given by:

E_initial = m * g * h_initial

Where:
m = mass of the skier
g = acceleration due to gravity (9.8 m/s^2)
h_initial = vertical distance at the edge of the cliff

The final mechanical energy of the skier is given by:

E_final = m * g * h_final + (1/2) * m * v^2

Where:
h_final = vertical distance below the edge (4.50 m)
v = speed at the edge of the cliff (what we're looking for)

Since there are no non-conservative forces acting on the skier (such as air resistance), the initial mechanical energy should be equal to the final mechanical energy:

E_initial = E_final

m * g * h_initial = m * g * h_final + (1/2) * m * v^2

Now, let's plug in the given values:

h_initial = 0 (since the skier starts from rest)
h_final = -4.50 m (negative because it is below the initial position)

m * g * 0 = m * g * (-4.50 m) + (1/2) * m * v^2

Simplifying the equation:

0 = -4.50 * 9.8 m/s^2 + (1/2) * v^2

Rearranging the equation:

(1/2) * v^2 = 4.50 * 9.8 m/s^2

v^2 = 2 * 4.50 * 9.8 m/s^2

v^2 = 88.2 m^2/s^2

Finally, taking the square root of both sides to find v:

v = √(88.2 m^2/s^2)

v ≈ 9.39 m/s

Therefore, the skier's speed just before landing is approximately 9.39 m/s.