Question

A 2 kg object released from rest at the top of a tall cliff reaches a terminal speed of 37.5 m/s after it has fallen a height of 110 m . How much kinetic energy approximately did the air molecules gain from the falling object after it has fallen through this height?

Answer 1

Answer:

3562.25 J

Explanation:

m = mass of the object = 2 kg

v = speed of the object after falling through height of 110 m , = 37.5 m/s

h = height through which the object fall = 110 m

KE = Kinetic energy gained by air molecules

Kinetic energy gained by air molecules is given as

KE = (0.5) m v² + mgh

KE = (0.5) (2) (37.5)² + (2) (9.8) (110)

KE = 3562.25 J