A flowerpot falls off the top of a building which is 70 meters high. If the flowerpot has a mass of 2.4 kg, what is its potential energy after 2.5 seconds?

potential-energy

#1

A flowerpot falls off the top of a building which is 70 meters high. If the flowerpot has a mass of 2.4 kg, what is its potential energy after 2.5 seconds? Keep 3 significant figures.

Answer:

Given,
Height of the building = 70 meter
Mass of flowerpot = 2.4 kg
Potential energy after 2.5 seconds of free fall = “?”

By Newton’s Second Law of motion
we have: s = ut + 1/2at^2
In this case,
u=0, t=2.5 sec., a = g(9.8m/s^2).
So the vertical distance travelled by the free falling flowerpot, in 2.5 seconds
= 1/29.82.52.5
=30.625 meter.
i.e. Height of flowerpot after 2.5 seconds = (70 - 30.625) meter
=39.375 meter
Therefore, potential energy after 2.5 seconds of free fall
= m
gh
=2.4
9.8*39.675 N-m
=926.100 N-m