Question

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0 with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 10.4 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 3.50 m below the edge. How fast is she going just before she lands?

Answer 1

Answer:

V = 10.88 m/s

Explanation:

V_i =initial velocity = 0m/s

a= acceleration= gsinθ-$\mu_k$cosθ

putting values we get

a= 9.8sin25-0.2cos25= 2.4 m/s^2

v_f= final velocity and d= displacement along the inclined plane = 10.4 m

using the equation

$v^2_f=v^2_i-2as$

$v^2_f=0^2-2(2.4)(10.4)$

v_f= 7.04 m/s

let the speed just before she lands be "V"

using conservation of energy

KE + PE at the edge of cliff = KE at bottom of cliff

(0.5) m V_f^2 + mgh = (0.5) m V^2

V^2 = V_f^2 + 2gh

V^2 = 7.04^2 + 2 x 9.8 x 3.5

V = 10.88 m/s

Answer 2

The speed of the skier just before she lands is 10.86 m/s.

Acceleration of the skier

The acceleration of the skier is determined from Newton's second law of motion as shown below;

∑F = ma

mgsinθ - μmgcosθ = ma

gsinθ - μgcosθ = a

g(sinθ - μcosθ) = a

9.8(sin25 - 0.2 x cos25) = a

2.37 m/s² = a

Final velocity at the edge of the cliff

v² = u² + 2as

v² = 0 + 2(2.37)(10.4)

v² = 49.296

v = √49.296

v = 7.02 m/s

Final velocity from the vertical distance

Apply principle of conservation of energy

ΔK.E = ΔP.E

¹/₂mvf² - ¹/₂mv² = mghf - mgh0

¹/₂vf² - ¹/₂v² = ghf

vf² - v² = 2ghf

vf² = 2ghf + v²

vf² = 2(9.8)(3.5) + (7.02)²

vf² = 117.88

vf = 10.86 m/s

Thus, the speed of the skier just before she lands is 10.86 m/s.

Learn more about conservation of energy here: https://brainly.com/question/24772394