The Relationship Between KE and PE

Kinetic energy is the energy that an object with mass will have when it is in motion.

The equation for kinetic energy is: KE=\cfrac { 1 }{ 2 } m{ v }^{ 2 }

Also remembering that gravitational potential energy: U=mgh

The law of conservation of energy states that energy cannot be created nor destroyed, but can be changed from one form to another. Following from this we can consider problems where an object near the Earth’s surface is dropped. When an object is dropped its velocity will increase from 0 and its height will decrease. This leads us to assume that the objects gravitational potential energy will decrease and its kinetic energy will increase. The total sum of these energies at any point will always be constant.

This relationship is Illustrated below. Consider a rock that is dropped from a cliff. At the top of the cliff the energy of the system is 100% { U }_{ p }. As the rock falls the PE decreases and the KE increases. This occurs until the rocks energy is 100% KE at h=0

Energy Changes Near the Earth’s Surface

Close to the surface of the Earth, we arbitrarily take the h=0 position as being ground level because it is much simpler to make calculations to predict velocities of falling objects this way.

For problems where an object with mass is dropped from a height:

Total energy, TE=KE+PE

TE=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+mgh

Any change in an objects potential energy as it falls will be converted to kinetic energy: \Delta { U }_{ p }=\Delta KE

Example 1: 

Calculate the KE of a 1000kg vehicle travelling at 60\cfrac { km }{ hr } :

Using: KE=\cfrac { 1 }{ 2 } m{ v }^{ 2 }

We must convert the 60\cfrac { km }{ hr } to \cfrac { m }{ s } :

60\cfrac { km }{ hr } \times \cfrac { 1000m }{ 3600s } =16.67\cfrac { m }{ s }

KE=\cfrac { 1 }{ 2 } \times 1000\times 16.67^{ 2 }


J=138.9\quad kJ

Example 2: 

An object with a mass of 2kg was dropped from the top of a 10m platform. Calculate the final velocity of the object using the conservation of energy law:

Gravitational potential energy; mgh  at the top of the platform will be converted to kinetic energy; \cfrac { 1 }{ 2 } m{ v }^{ 2 } at the bottom.

 U_{ p }=mgh=2\times 9.8\times 10

 U_{ p }=196\quad J

As all of this energy is converted to KE:

196=\cfrac { 1 }{ 2 } m{ v }^{ 2 }

196=\cfrac { 1 }{ 2 } \times 2\times { v }^{ 2 }

{ v }^{ 2 }=196

{ v }=14\cfrac { m }{ s }


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