**Content – Forms of energy**

**Kinetic energy is the energy of an object caused by its motion.**

**The kinetic energy is a function of the mass and the velocity of the object.**

The kinetic energy of an object equals the work necessary to accelerate the object to the actual speed from still stand, disregarding any work needed to overcome friction.

For most objects the total kinetic energy (E_{k}) consist of two main components; the translational kinetic energy (E_{t}) which simplified can be described as the kinetic energy of the centre of mass of the object and; the rotational kinetic energy (E_{r}) the energy of rotation around the centre of mass.

*Total Kinetic energy*

*Total Kinetic energy*

E_{k} = E_{t}+ E_{r}

E_{t} = Translational kinetic energy

E_{r} = Rotational kinetic energy

**Formula for translational kinetic energy**

E_k=\dfrac{1}{2}mv^2

m = mass (Kg)

v = velocity (m/s metres per second)

E_{k }= Resulting energy is measured in joules

**The use of the above “classical” formula for kinetic energy assumes that the speed of the object is less (much less) than the speed of light (“c”).**

If the speed of the object is approaching the speed of light the formula for relativistic kinetic energy needs to be used.

The formula for relativistic kinetic energy may be found here: Formula for relativistic kinetic energy.

**Formula for rotational kinetic energy**

E_r=\dfrac{1}{2}Iw^2

*I* = Moment of inertia (around the axis of rotation)

ω = Angular velocity = 2πf

f = Revolutions/sec

**For more energy formulas go to: Formulas – energy**

*Example – calculating kinetic **energy*

*Example – calculating kinetic*

The kinetic energy of a motorbike with driver with a combined mass of 350 kg and a speed of 16 m/s (60 Km/hr.) would be (disregarding any rotational kinetic energy of the wheels and the machinery):

E_k=\dfrac{1}{2}350kg*(16m/s)^2 = 44800 J = 44,8 KJ

This equals the amount of work needed to accellerate (disregarding friction) or stop the motorbike from the speed of 60 km/hr.

To stop another motorbike MC_{2} (same weight of bike and driver) that runs at the double speed (120 km/hr) it would requier **4 times the energy** of the first motorbike MC_{1}, demonstrated by the following calculation:

MC _{2}, runs at 120 Km/hr. (32 m/s)

E_k=\dfrac{1}{2}350kg*(32m/s)^2 = 179200 J = 179,2 KJ

MC_{2}/MC_{1} = 179,2/44,8 = 4* *