In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,
Where,
M = Mass of Earth
m = Mass of Object
v = Velocity
r = Radius
G = Gravitational universal constant
Our values are given as,
Replacing we have,
Therefore the speed of the object when striking the surface of earth is 4456 m/s
Answer:
The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
Explanation:
It is given that,
Initial gravitational potential energy of the ball-Earth-floor system is 10 J.
The ball then bounces back up to a height where the gravitational potential energy is 7 J.
Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.
Initial mechanical energy is 10 + 0 = 10 J
Mechanical energy just before the collision is 0 + 10 = 10 J
Final mechanical energy, 7 + 0 = 7 J
Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
