A 25 kg bear slides, from rest, 8m down a lodgepole pine tree, moving with a speed of 5.7 m/s just before
hitting the ground. What is the average frictional force that acts on the bear?
2006-10-30 11:41:05 · 2 answers · asked by Shane H 2 in Science & Mathematics ➔ Physics
Consider the energy lost to friction
E = Fd = mgd - mv^2/2
from which
F = mg - mv^2/(2d)
the rest is arithmetic
2006-10-30 16:18:19 · answer #1 · answered by d/dx+d/dy+d/dz 6 · 0⤊ 0⤋
It's been a while since I took physics, and I'm not sure of the formulas. The approach I would take is as follows. Calculate the bears acceleration as he slides down the pole. Acceleration due to gravity should be 9.81m/s^2. The difference results from friction. The amount of force required to accelerate the bear at the rate his actual acceleration differs from gravity is the frictional force.
2006-10-30 12:47:00 · answer #2 · answered by STEVEN F 7 · 0⤊ 0⤋