Question

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Answer 1

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

           = $0.5 x 25 x 5.6^{2}$  = 392 J

(C) average frictional force = $\frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}$

• change in KE (ΔKE) = initial KE - final KE
• ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$            
• when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

=  $\frac{(-392 + 2940)}{12}$ = 212.33 N