Question

Starting from rest near the surface of the Earth, a 25-kg beam slides 12 m down a vertical pine tree, and has a speed of 6 m/s just before hitting the ground. What is the magnitude of the average frictional force (in N) acting on the bear during the slide

Answer 1

The frictional force acting on the bear during the slide is 207.5 N

Step-by-step explanation:

Given;

mass of beam, m = 25-kg

vertical height, h = 12 m

speed of fall, v = 6 m/s

Change in potential energy of the beam:

ΔP.E = -mgh = - 25 x 9.8 x 12 = -2940 J

Change in kinetic energy of the beam:

Δ K.E = ¹/₂mv² = ¹/₂ x 25 x (6)² = 450 J

Change in thermal energy of the system due to friction:

ΔE = - (ΔP.E + Δ K.E)

ΔE = - (-2940 J + 450 J)

ΔE = 2940 J - 450 J = 2490 J

Frictional force (in N) acting on the bear during the slide:

F x d = Fk x h = ΔE

Where;

Fk is the frictional force

Fk = ΔE/h

Fk = 2490J / 12m

Fk = 207.5 N

Therefore, the frictional force acting on the bear during the slide is 207.5 N