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 March 15th, 2017, 11:20 PM #1 Newbie   Joined: Feb 2017 From: Lebanon Posts: 18 Thanks: 1 Why does kinetic energy increase as velocity squared? The formula for kinetic energy is E=[1/2]mv^2 and the formula for momentum is P=mv. I ran into these in physics class long ago and was really bothered by the first formula. How can energy go up as the square of the velocity?
 March 16th, 2017, 12:04 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 suppose we throw a ball vertically into the air with velocity $v_0$ At that point in time the ball has a certain amount of kinetic energy and as it rises against the force of gravity it does work against the gravitational field until it has exhausted it's supply of kinetic energy and stops at some height $h$. We know that in the absence of friction the ball has converted it's kinetic energy into gravitational potential energy. The potential energy gain is $m g h$ the balls stops moving at time $t = \dfrac{v_0}{g}$ and the ball attains a height of $h = v_0 t - \dfrac{g t^2}{2} = \dfrac{v_0^2}{g} - \dfrac{g v_0^2}{2 g^2} = \dfrac 1 g \left(v_0^2-\dfrac{v_0^2}{2}\right) = \dfrac{v_0^2}{2g}$ so the gravitational potential gained, and hence the kinetic energy started with is $\Delta U = KE = m g \dfrac{v_0^2}{2g} = \dfrac 1 2 m v_0^2$ You can also do a units analysis. Energy is work which is force times distance Force is mass times acceleration so using fundamental units, M, L, T, for mass, length, and time we have $E \sim M \dfrac{L}{T^2} L = M \left(\dfrac{L^2}{T^2}\right) = M \left(\dfrac{L}{T}\right)^2$ $\dfrac{L}{T}$ corresponds to velocity so again we have $E \sim M V^2$ Thanks from topsquark, v8archie, Benit13 and 1 others
 March 16th, 2017, 11:19 AM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,256 Thanks: 926 Math Focus: Wibbly wobbly timey-wimey stuff. romsek has given you a very good answer. But I thought I'd add my two cents anyway. Energy is a very hard concept to pin down. There are a number of quantities in Physics that have that "property." Momentum, energy, force, and mass are some others that are difficult to pin down as well. So why do we define them the way we do? Simple: Over time we find that some definitions are more useful than others. The way we define energy has a nice property...it is conserved for closed systems. So basically the definition for kinetic energy is as it is because it's more useful than any other way to do it. A lot of definitions in Physics are made that way. -Dan Thanks from romsek, JeffM1 and JamSmith
 March 16th, 2017, 04:06 PM #4 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 for uniform acceleration ... $v_f^2 = v_0^2 + 2a \Delta x$ $v_f^2 - v_0^2 = 2a \Delta x$ $\dfrac{v_f^2}{2} - \dfrac{v_0^2}{2} = a \Delta x$ $m\bigg[\dfrac{v_f^2}{2} - \dfrac{v_0^2}{2}\bigg] = ma \Delta x$ $\dfrac{mv_f^2}{2} - \dfrac{mv_0^2}{2} = F_{net} \Delta x$ $\Delta KE = W_{net}$ Thanks from topsquark and JamSmith
 March 20th, 2017, 05:06 AM #5 Newbie   Joined: Feb 2017 From: Lebanon Posts: 18 Thanks: 1 Thank yo all for the calculation. but, I have another confusion:- So Why does kinetic energy only increase quadratically and not linearly, with speed?
March 20th, 2017, 06:20 AM   #6
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Quote:
 Originally Posted by JamSmith Thank yo all for the calculation. but, I have another confusion:- So Why does kinetic energy only increase quadratically and not linearly, with speed?
It's just defined that way. Momentum varies linearly with speed and is a different thing to energy.

March 22nd, 2017, 04:21 AM   #7
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Why does kinetic energy increase as velocity squared?

Quote:
 Originally Posted by Benit13 It's just defined that way. Momentum varies linearly with speed and is a different thing to energy.
So what is the difference between Momentum & Kinetic energy? As you said, kinetic energy can be converted into other forms of energy. But, I have read somewhere that Kinetic energy in itself is not necessarily obtained only the total energy.

March 22nd, 2017, 06:12 AM   #8
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Quote:
 Originally Posted by JamSmith So what is the difference between Momentum & Kinetic energy? As you said, kinetic energy can be converted into other forms of energy. But, I have read somewhere that Kinetic energy in itself is not necessarily obtained only the total energy.
Some differences are

1) Momentum is a vector, kinetic energy is a scalar, like all forms of energy.

2) Momentum is always a conserved quantity Kinetic Energy is not. I think your reference from elsewhere is to total mechanical energy in a purely mechanical system which is a conserved quantity.

3) There is only one form of momentum and this cannot be converted into anything else, unlike energy which takes many different forms. It is these different forms that are the reason why Kinetic Energy alone is not conserved.

 June 18th, 2019, 12:55 PM #9 Newbie   Joined: Jun 2019 From: United States Posts: 1 Thanks: 0 The simplest way to look at this is to consider how the mass, momentum and kinetic energy transform under a change to second frame of reference moving with respect to the first one. Let m, p and H denote the mass, momentum and kinetic energy of a body, the quantity p being a vector. Upon transformation to a second system moving with an infinitesimal velocity du with respect to the first system, neglecting terms of the second order in du, since the underlying velocities v of everyone will be changed by dv = -du, after being referred to the second system, then these quantities will undergo a change by dm = 0, dp = -m du ... and dH = -p·du. Recall that the kinetic energy accumulates the total work done on the body and that the work done by a force F takes place at a rate given by F·v, in the first frame. In the second frame, that rate is altered by -F·du, and, so likewise the kinetic energy. Recalling also that the force impressed on a body is equal to the rate of change of its momentum, then yields the transformation law for the kinetic energy, as stated. Already this makes it clear that everything will be quadratic, rather than linear: the impact on H will go as the square of the change du. The transform affects the mass, momentum and kinetic energy respectively as the 0, 1st and 2nd power of du. From the transformation under an infinitesimal change du, you can infer the transformation under a finite change u, by integrating from 0 to u, with the following results:m → m, p → p - ∫ m(u) du = p - ∫ m du = p - m ∫du = p - mu, H → H - ∫ p(u)·du = H - ∫ (p - mu)·du = H - (p·∫du - m∫u·du) = H - pu + ½mu² So, finally, consider what happens when you transform to the rest frame of the body itself. In that frame, the momentum is 0; while, by definition, the momentum of the body in the moving frame is given in terms of its velocity by p = mv. Supposing at the kinetic energy of the body is just its internal energy U (the total internal kinetic energy of all of the parts that make up the body), then the resulting transform is: m = m, 0 = mv - mu, U = H - mv·u + ½mu²which implies u = v, as it should, and leads the formulae p = mv and H = ½mv² + U.

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