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]
\frac{-(ΔKE+ΔU)}{h}[/tex] = [tex]\frac{-(392 + (-2940))}{12}[/tex]
= [tex]\frac{(-392 + 2940)}{12}[/tex] = 212.33 N
(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.
The gravitational potential energy of the bear decreases as the bear slides from top to bottom of the pine tree.
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
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
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