43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 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?

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

The frictional force that acts on the sliding bear is 378.47166 N

Answer

mass of bear = 43 Kg                  

length of the pine tree = 15 m

speed = 5.5 m/s            

a) P.E = m g h                

P.E = 43 x 9.8 x 15

P. E = 6321 J                

b)

c) average frictional force              

K E = P E + energy loss due to friction

650.375 = 6321 + E_f

E_f = -5670.625 J

E_f = F x                              

-5670.625 = F x 15

F = -378.042 N

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

the components of a vector are the parts of the vectors that are to each other.

b perpendicular