Gravitational potential energy and work done

If an object is lifted, work is done against the force of gravity.

When work is done energy is transferred to the object and it gains gravitational potential energy.

If the object falls from that height, the same amount of work would have to be done by the force of gravity to bring it back to the Earth’s surface.

If an object at a certain height has 2000 J of gravitational potential energy, we can say that:

2000 J of work has been done in getting the object to that height from the ground

and

2000 J of work would have to be done to bring it back to the ground.

curriculum-key-fact
Gravitational potential energy = work done

Kinetic energy, gravitational potential energy and conservation of energy

If an object, such as a ball is lifted above the ground it has gravitational potential energy.

If the ball is then dropped from rest it will fall back to the ground. The gravitational potential energy is converted to kinetic energy.

Due to the Principle of Conservation of Energy we can say that:

Gravitational potential energy at the top = kinetic energy at the bottom

GPEtop = KEbottom

Question

A ball of mass 0.4 kg is lifted to a height of 2.5 m.

It is then dropped, from rest.

What is the speed of the ball as it hits the ground (g = 10 N/kg)

GPEtop = KEbottom

m = 0.4 kg

g = 10 N/kg

h = 2.5 m

(mgh)top = ( \frac{1}{2} mv2)bottom

0.4 kg x 10 N/kg x 2.5 m = \frac{1}{2} x 0.4kg x v2

10 = 0.2 v2

v2 = \frac{10}{0.2}

v2 = 50

v = \sqrt{50}

v = 7.1 m/s

The ball returns to the ground with a speed of 7.1 m/s.

Question
  1. A mass of 25 kg is dropped from the top of a tower 20 m high. What is the speed of the mass as it hits the ground?
  2. A mass of 50 kg is then dropped from the same height. What is its speed as it hits the ground?

1. GPEtop = Kebottom

m = 25 kg

g = 10 N/kg

h = 20 m

(mgh)top = ( \frac{1}{2}mv2)bottom

25 kg x 10 N/kg x 20 m = \frac {1}{2} x 25 kg x v2

5000 = 12.5 v2

v2 = \frac{5000}{12.5}

v2 = 400

v = \sqrt{400}

v = 20 m/s

The 25 kg mass returns to the ground with a speed of 20 m/s.

2. GPEtop = Kebottom

m = 50 kg

g = 10 N/kg

h = 20 m

(mgh)top = ( \frac{1}{2}mv2)bottom

50 kg x 10 N/kg x 20 m = \frac {1}{2} x 50 kg x v2

10 000 = 25 v2

v2 = \frac{10,000}{25}

v2 = 400

v = \sqrt{400}

v = 20 m/s

The 50 kg mass returns to the ground with a speed of 20 m/s, which is exactly the same as the speed of the 25 kg mass.

If there was no air resistance or drag, the 25 kg mass and the 50 kg would fall at the same rate of 10 m/s2.

Dropped from the same height, they both hit the ground at the same speed and after the same period of time.

This is true of all objects regardless of their mass - in the absence of air resistance (friction) they fall at the same rate.