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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.
(a) What change occurs in the gravitational potential energy of the bear – Earth system during the slide?
(b) What is the kinetic energy of the bear just before hitting the ground?
(c) What is the average frictional force that acts on the sliding bear?

Respuesta :

usmanbiu

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) = [tex]0.5mv^{2}[/tex]

           = [tex]0.5 x 25 x 5.6^{2}[/tex]  = 392 J

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

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

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

=  [tex]\frac{(-392 + 2940)}{12}[/tex] = 212.33 N

onyebuchinnaji

(1) The gravitational potential energy of the bear decreases as the bear slides from top to bottom of the pine tree

(2) The kinetic energy of the bear just before hitting the ground is 392 J.

(3) The average frictional force that acts on the sliding bear is 212.33 N.

Change in the gravitational potential energy

The gravitational potential energy of the bear decreases as the bear slides from top to bottom of the pine tree.

Kinetic energy of the bear hitting the ground

The kinetic energy of the bear before hitting the ground is calculated as follows;

K.E = 0.5 mv²

K.E = 0.5 x 25 x 5.6²

K.E = 392 J

Average frictional force acting on the bear

The average frictional force acting on the sliding bear is determined from the change in mechanical energy of the bear.

ΔE = P.E - K.E

ΔE = mgh - K.E

ΔE = (25 x 9.8 x 12) - 392

ΔE = 2940 - 392

ΔE = 2548 J

F x d = ΔE

F = ΔE /d

F = 2548/12

F = 212.33 N

Learn more about kinetic energy here: https://brainly.com/question/25959744

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