21 kg bear slides, from rest, 12m down a lodgepole pine tree, moving at a speed of 4.5 m/s just before land?

Question

) What is the average frictional force that acts on the bear?
Physics Homework

Answer 1

Use Torricelli's equation to find the acceleration down the pole:
vf^2 = vi^2 + 2ad
(4.5 m/s)^2 = 0 + 2a * 12
20.25 = 24a
a = 0.84375 m/s^2

Since F = ma and is the difference between the gravity pulling the bear and the frictional force, we can rewrite it to:
F = Fg - Ff
ma = mg - Ff
ma = 21 * 0.84375 = 17.71875
mg = 21 * 9.8 = 205.8
Ff = mg - ma
Ff = 205.8 - 17.7 = 188.1 N