Answer:
B. A object in motion stays in motion, and an object at rest stays at rest unless acted upon by a net force.
Answer: 12 m/s
Explanation:
If we ignore air resistance, gravity alone will reduce the upward velocity to zero at the top of the flight. As gravity is a conservative force, it will return exactly the same amount of energy to the tennis ball when it returns to the original elevation.
1) KE=1/2*m*v^2
1/2*45*40^2
KE=36,000J
2) PE=mgh
45*9.81*30
PE=13243.5J
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
