Question

A hot air balloon drops a 20 kg package from rest at a height of 120 m from the ground. Assume we can ignore friction losses, as the package falls through the air.

a) How much initial potential energy is stored by the package?

b) What is the kinetic energy of the package just before it hits the ground?

c) What is the speed of the energy before it hits the ground?

Answer 1

Given : M= 20 kg , h = 120 m

constant : Acceleration due to gravity (g)= 9.81 m/s²

Solution:

(a) Initial stored potential energy

The initial potential energy stored by the package is given by:

PE = Mgh

= (20 kg)(9.81 m/s²)(120 m)

= 23544 J

Answer : 23544 J

(b) Kinetic energy of package just before ground

According to the conservation of energy , all of its initial stored potential energy will be transformed into kinetic energy.

i.e. the kinetic energy = potential energy

So, kinetic energy = 23544 J

Answer : 23544 J

(c) Speed of the package just before it hits the ground

As we have determined , kinetic energy = 23544 J

We know that , kinetic energy is given by: KE = (1/2)Mv² where v is the velocity of the package.

i.e. (1/2)Mv² = 23544 J

(1/2)(20)v² = 23544

v² = 2354.4

v = 2354.4 = 48.52 m/s

Answer : 48.52 m/s

answered by: anonymous