Question

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?

Answer 1

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) $K.E = \dfrac{1}{2} mv^2$

$K.E = \dfrac{1}{2}\times 43 \times 5.5^2$

$K.E = 650.375\ J$

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

Answer 2

Answer:

-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

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

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

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

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

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$

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