Answer:
A. -5488J
B. 273.8J
C. 372.44N
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
Answer:
Explanation:
(a) Here, we have to take into account the earth's radius and the value of the Cavendish constant
[tex]E_{p1}-E_{p2}=G\frac{m_{b}m_{e}}{r'}-G\frac{m_{b}m_{e}}{r}=\\6.67*10^{-11}\frac{Nm^{2}}{kg^{2}}\frac{40kg*5.9*10^{24}kg}{6371*10^{3}m}-6.67*10^{-11}\frac{Nm^{2}}{kg^{2}}\frac{40kg*5.9*10^{24}kg}{6371*10^{3}m+14m}=\\0.37J[/tex]
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). [tex]E_{k}=\frac{mv^{2}}{2}=\frac{(40kg)(3.7m/s)^{2}}{2}=273.8J[/tex]
(c). Is the force produced principally by the tree
[tex]F_{f}=-\mu N[/tex]
I hope this is usefull for you
