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QUESTION: Hi,

I've always wondered why kinetic energy's mv^2 is divided by two.

Below are two answers that I've gotten, could you please critique them and let me know anything true or false in them and which one is the right answer if either. And if both are wrong what the real answer is.

Thanks,

Felicity

Dear Felicity,

For any constantly accelerated movement you can draw a graph of the change of velocity against the path traveled. At some arbitrary initial point the velocity change is zero. As the acceleration is constant, the velocity increases linearly with the path traveled, up to an arbitrary end point where the velocity change is maximal. This graph forms a right

angle triangle, and you can draw it along any infinitesimal part of the orbit. Now we should ask ourselves what is a constant velocity by which the object would travel the same path in the same amount of time. It is obvious that it is equal to 1/2 of the maximum. That is where the factor 1/2 comes from.

Hi Felicity,

When Gravesande dropped the lead ball into the soft clay, the ball started from rest and accelerated until it hit the clay. That actual falling "Distance" that the ball travels is only "half" the distance of (Acceleration x Time^2). Therefore that is the reason the kinetic energy equation of a falling object is divided by 2. Hence 1/2 mv^2. However with an orbiting planet this is not so. The Velocity of the planet (Acceleration x Time) multiplied by the orbit Time is exactly comparable to the "Falling Distance" of Gravesande's ball. The planet is not starting from rest therefore you would not divide the mv^2 in half. The planet is traveling at "final velocity" throughout the orbit therefore if you divide the mv^2 in half, you are robbing it of half of it's true energy. Therefore the kinetic energy of an orbiting planet = mv^2 not 1/2 mv^2

ANSWER: Here is a simple example.

A ball is located on the top of a building h meters high. As a result the ball has gravitational energy equal to:

GPE=m*g*h

The displacement of a uniformly accelerated object is given by:

D=1/2*a*t^2

If you drop that ball to the ground below the position of the ball as a function is given by:

h=1/2*g*t^2 (1)

Where g is the magnitude of the gravitational acceleration.

Likewise, for a uniformly accelerating object the velocity from rest is given by:

v=a*t (2)

Which for the ball dropped from the top of the building becomes:

v=g*t

If you multiply each side of equation (1) above by the mass m and the gravitational acceleration g it becomes:

h=1/2*g*t^2

m*g*h=1/2*m*g^2*t^2=1/2*m*(g*t)^2

Substituting v, equation (2), in for g*t:

m*g*h=1/2*m*v^2

Since the left side of this equation is energy (gravitational potential energy) the right side must also be energy! But this energy on the right side is written in terms of velocity v rather than height h and so is kinetic energy!

GPEo=KEf

KE=1/2*m*v^2

---------- FOLLOW-UP ----------

QUESTION: Thanks James I really appreciate that.

What about a planet orbiting ?

It's displacement is a*t^2 not 1/2 at^2

Therefore its KE = mv^2 not 1/2 mv^2

Right ?

I have no idea where you got the idea that the displacement of an orbiting satellite would be a*t^2. That is just not correct. For a satellite moving in a circular orbit the distance moved would be equal to the speed of the satellite in its orbit multiplied by time, certainly not time squared.

Physics

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.