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

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

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

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle