2.2c – Calculating energy in the kinetic store

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The an object's kinetic store of energy can be calculated using the equation:

cap e times n times e times r times g times y sub k times i times n times e times t times i times c s times t times o times r times e equals one divided by two multiplication m times a times s times s multiplication v times e times l times o times c times i times t times y squared

cap e sub cap k equals one divided by two times m times v squared

The SI unit for energy is the joule (abbreviated as cap j ).  Mass is measured in k times g , and velocity in m solidus s .

Example kinetic store question

Train with force F over distance sA locomotive of mass 210 000 k times g is travelling at 40 m solidus s , how much energy is stored kinetically?

equation sequence part 1 cap e sub cap k equals part 2 one divided by two times m times v squared equals part 3 one divided by two multiplication 210000 multiplication 40 squared equals part 4 168 000 000 cap j equals 168 cap m times cap j

Deriving the kinetic store equation

The energy stored kinetically equation seems to suddenly appear. Although your students do not need to know where this equation comes from, it would be useful for you as a teacher to see that it originates from the equation for work done.

cap w equals cap f multiplication s

Let’s use a simple example of a train accelerating on a track. We will assume that all the energy is shifting from the chemical store in the diesel to the kinetic store of the train. In other words, we will ignore drag forces so there are no energy transfers into the thermal stores of the train, ground or surroundings.

The energy shifted mechanically from the chemical store to the kinetic store can be expressed as

cap w equals cap f times s

But we know from Newton’s Second Law:

cap f equals m times a

where m is the mass of the train and a is the acceleration. So, we can express the work done as:

cap w equals m times a times s

But we have also seen from the Motion section in the Forces module (link) that:

v squared equals u squared plus two times a times s

where v is the final velocity of the train after travelling a distance s and u is the initial velocity. So, if we rearrange and divide both sides by 2 then

a times s equals one divided by two times left parenthesis v squared minus u squared right parenthesis

And our equation for work done becomes:

cap w equals one divided by two times m times left parenthesis v squared minus u squared right parenthesis

If the initial velocity u of our train is zero, this simplifies to:

w times o times r times k d times o times n times e equals one divided by two times m times v squared

cap e sub k times i times n times e times t times i times c equals one divided by two times m times v squared

This is the equation for the energy stored kinetically by the train.

2.2b – Calculating energy in the gravitational store

2.2d – Calculating energy in the elastic store

Last modified: Friday, 7 January 2022, 4:34 PM