Question

A snowboarder starts from rest on an inclined planar snow surface and gives herself a tiny push to start sliding down the hill. The coefficient of kinetic friction between her snowboard and the snow is 0.14. She notices that she travels 27.81 m horizontally and 39.78 m vertically (downward) before whizzing by a marker flag. How fast is she traveling as she passes the flag?

Answer 1

Answer:

v = 26.52 m/s

Explanation:

Here we know that as she passes the flag she will gain some speed

so here we can say that

Work done by all forces = change in kinetic energy of the system

so we will have

$W_g + W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

here we know that

$W_g = mgh$

$W_f = -\mu mg d$

$v_i = 0$

h = 39.78 m

d = 27.81 m

so we have

$mgh - \mu mg d = \frac{1}{2}mv^2 - 0$

$(9.8)(39.78 - (0.14)(27.81)) = \frac{1}{2} v^2$

$v = 26.52 m/s$