Question

A 40 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 3.7 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

Step-by-step explanation:

(a) Here, we have to take into account the earth's radius and the value of the Cavendish constant

`E_(p1)-E_(p2)=G(m_(b)m_(e))/(r')-G(m_(b)m_(e))/(r)=\\6.67*10^(-11)(Nm^(2))/(kg^(2))(40kg*5.9*10^(24)kg)/(6371*10^(3)m)-6.67*10^(-11)(Nm^(2))/(kg^(2))(40kg*5.9*10^(24)kg)/(6371*10^(3)m+14m)=\\0.37J`

The change is very low due to we consider the radius of the earth as the distance to the center of mass of the earth

(b).
`E_(k)=(mv^(2))/(2)=((40kg)(3.7m/s)^(2))/(2)=273.8J`

(c). Is the force produced principally by the tree

`F_(f)=-\mu N`

I hope this is usefull for you

Answer 2

A. -5488J

B. 273.8J

C. 372.44N

Step-by-step explanation:

Given:

m = 40kg

h = 14 m

v= 3.7 m/s

Part(a)

The change in the potential energy of the bear Earth system during the slide

AU = -mgh = -40(9.8) (14) = -5488 J

Part(b)

The kinetic energy of the bear just before hitting the ground is

Ks 1/2 mV^2= (40)(3.7)2 = 547.6 /2 = 273.8J

Part(c)

The change in the thermal energy of the system due to friction is

AEth = fxh=-(AK +AU) = 5488– 273.8 = 5214.2 J

The average frictional force that acts on the sliding bear is

F = Eth / 14= 5214.2/14 =372.44N