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