Question

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy even on small hills.) To demonstrate this, find the final speed in m/s and the time taken in seconds for a skier who skies 64.0 m along a 28° slope neglecting friction for the following two cases. (Enter the final speeds to at least one decimal place.)

(a) starting from rest

final speed in m/s

time taken in seconds

(b) starting with an initial speed of 3.50 m/s

final speed in m/s

time taken in s

Answer 1

(a) we know that, sin 28⁰ = h / (64 m)

h = 30 m

using conservation of energy, we have

K.E_i + P.E_i = K.E_f + P.E_f

(1/2) m v_i² + m g h = (1/2) m v_f² + (0 J)

where, v_i = starting from rest = 0 m/s

THEN, we get

v_f = 2 g h = 2 (9.8 m/s²) (30 m)

v_f = 24.2 m/s

We know that, average velocity = (v + v_i) / 2

Time taken is given by, t = x / v_avg = (64 m) / [(24.2 m/s) / 2]

t = 5.28 sec

(b) Again, using conservation of energy -

K.E_i + P.E_i = K.E_f + P.E_f

(1/2) m v_i² + m g h = (1/2) m v_f² + (0 J)

where, v_i = starting with an initial speed = 3.5 m/s

THEN, we get

v_f² = 2 g h + v_i² = 2 (9.8 m/s²) (30 m) + (3.5 m/s)²

v_f = 600.2 m²/s²

v_f = 24.5 m/s

We know that, average velocity = (v + v_i) / 2

Time taken is given by, t = x / v_avg = (64 m) / [(24.5 m/s) + (3.5 m/s) / 2]

t = 4.57 sec