Question

An eagle that has a mass of 4.5 kg is spotted 5 m above a lake

Answer 1

An Eagle that has a mass of 4.5 kg is spotted 5 m above a lake. It flies up to its nest in a tree at a height of 25 m. How much gravitational potential energy did the eagle gain?

This is the original question

Answer:

882 J = PEg = MGH M = mass in Kg G = 9.6 m/s H = height

Height in this case is the displacement (25m - 5m) = 20m

Hope this helps.

Answer 2

I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
PLEASE SEE ATTACHED IMAGE.

By definition, the potential energy is given by:
Ep = m * g * h
Where,
m: mass
g: gravity
h: height.
The potential energy change is:
Delta Ep = Epf- Epi
Delta Ep = m * g * hf - m * g * hi
Delta Ep = m * g * (hf - hi)
Substituting values:
Delta Ep = (4.5) * (9.8) * (25 - 5)
Delta Ep = 882 J
Answer:
the eagle gain:
Delta Ep = 882 J