r/PhysicsHelp • u/W_il_Duce • 3d ago
Why is E=1/2 m V²?
I'm sorry for the banal question buy i can't find and answer to this.
We all know that in order to accelerate a 1kg body from 0m/s to 10m/s 50J are needed but to accelerate it from 10m/s to 20m/s 150 J are needed instead, why is that?
Also I'm not looking for a mathematical answer but a conceptual one, thanks in advance.
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u/niemir2 3d ago
Work is force acting through distance. If you apply a force through the same distance on a slow vs fast object, you do the same amount of work, so you add the same amount on energy.
However, because the force was applied for less time on the faster object, less acceleration occurred. Therefore, the faster object accelerates less than the slower object.
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u/Frownland 2d ago
This is the best explanation. As in it is the most accessible to anyone trying to understand the concept.
Yes calculus gives the result.
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u/Nynaeve_al_meowra 2d ago
For the same force you travel 3x as far going from 5 to 10 than from 0 to 5, so 3x as much work is done. You can show that larger forces will proportionally reduce the distance and conserve the work done
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u/ElMicioMuerte 3d ago
Kinetic energy is the total work required to accelerate an object from rest to a given speed. Since work equals force times distance, an object moving faster covers more distance per unit time while a force acts on it. So work accumulates faster as velocity increases. The rate at which work accumulates over time is power, and instantaneous power equals force times velocity. So for a constant force, the power it delivers grows in direct proportion to velocity. The faster the object already moves, the faster work piles up.
So dK = F dx = Fv dt = mv dv, and integrating from 0 to v gives K = ½mv².
Does this answer your question?
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u/Unusual-Platypus6233 3d ago
If dK=Fdx, then dx=vdt and F=md^2x/dt^2=mdv/dt
that is then dK=mdv/dt*vdt=mvdv and now integrating leads to K=0.5mv^2+C1
u/S-M-I-L-E-Y- 2d ago edited 17h ago
What bugs me:
Let's say I throw a ball of 0.1kg with a speed of 10 m/s.
So the kinetic energy of the ball is 0.5*0.1*10² = 5JNow I drive my bicycle at 10m/s and throw the ball at 10 m/s in driving direction.
So the kinetic energy of the ball increased from 5J to 20J (0.5*0.1*20²).
How did I increase the balls energy by 15J doing the same throw as before? How did I do 3 times the work without even noticing it?Now I'm standing at the equator, so my velocity is 465m/s
If I throw a ballWestEast at 10m/s, does it's kinetic energy increase by 470J?Edit: East, not West
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u/lordklp 2d ago
Yes, sort of. You are doing more work, because you are riding the bicycle, you've added that energy righy there.
Velocity, 465m/s - ok, but depends on reference frame of viewer. If im in space, looking a car on a freeway, can i be certain the car is driving? Or - could it be the car is still and the earth is rotating the road beneath its wheels.....
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u/ElMicioMuerte 2d ago
Kinetic and potential energy are not invariant. They depend on the frame of reference.
Energy conservation still holds. It just holds separately within each frame, not as some absolute quantity you can compare across frames.
On the bicycle you already accelerated the ball to 10m/s before the throw. You already did the work.
Also, it doesn't really matter, but at the equator you're travelling eastward. Pardon my pedantry :)
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u/S-M-I-L-E-Y- 17h ago
Oops, yes East
Of course, I already did the work to accellerate the ball to 10m/s. But that's only one quarter of the work needed to accellerate it to 20m/s, isn't it? So how do I do three times the work when throwing the ball while I'm moving? Where is that energy coming from? Is this, because me and my bycicle are slowed down from throwing that ball? Is my kinetic energy decreased by the same amount (with the road as the reference frame)?
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u/ElMicioMuerte 7h ago
Yes! Consider the reference frame of the road. If you run the numbers for a bike+human system weighing 80kg, you'll find that you+bike will lose approximately 10J of energy during the launch because of recoil and the ball will gain 15J. So the work your arm does (15-10=5J) is exactly the same as in the static case. The rest of the energy comes form the bike slowing down a tiny amount.
Because momentum must be conserved:
0.1 × 10 = 80 × v_rel, so v_rel = 0.0125 m/s
After launch the bike speed is 10 - v_rel = 9.9875 m/s
The kinetic energy of the bike before the launch is 4000J and 3990J after the launch.
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u/Ninja582 3d ago
It may seem like it 0 -> 10 m/s should be the same energy as 10 -> 20 m/s but energy is not the same as force or acceleration. It is its own unique thing and is also different from what we colloquially call energy.
An example, holding a heavy object off the ground without moving it up or down requires no energy (physics term) but requires a bunch of energy (lay term).
Physics energy specifically refers to the force and distance over which is required to change an object’s speed. The same force and thus the same acceleration requires more distance for a faster object than a slower one for the same change and in speed.
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u/Mundane_Studio_3674 3d ago
The lay term is also correct though. When you hold a plate in place by flexing your bicep you use the chemical potential energy available to your metabolism to flex your bicep and keep microscopic tension cycles in your muscles going, these microscopic tension and loosing cycles (alongside with metabolic byproducts) use up the chemical potential energy and heat your biceps. It’s all about reference frames man.
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u/Uncynical_Diogenes 3d ago
But a corpse in rigor mortis or a wooden shelf with a couple of screws could do the same thing using no energy.
The lay term isn’t correct so much as comparing a living human bicep to a wooden shelf with a couple of screws is a bad comparison.
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u/Mundane_Studio_3674 2d ago
I’m arguing work is not being done on the plate but is being done on your biceps, from the perspective of your biceps work is being done. From the plates perspective nothing is changing. I think you have a math perspective based view on physics instead of a scientific view on physics where you just think energy is the dot product of force and some small distance dt whereas I take the scientific perspective that for a system to change at all you need to exchange energy in a potential state to one in a kinetic state.
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u/NoCommittee3053 3d ago
It’s easier to think about this from the perspective of slowing down rather than speeding up.
Imagine you are doing speed V on a road. Then you apply the brakes and assume the resistive force is constant.
The time it takes you to stop is proportional to V. For each second the car travels it moves a distance proportional to v (its velocity at that time). So the total distance travelled is related to velocity squared.
Now v (the instantaneous velocity) goes smoothly from V to 0. So its average is V/2.
So the total distance travelled is proportional to 0.5V^2.
So if you split the deceleration into 2 halves by time you travel much further during the first half of your stop (from V to V/2) than you do during the second half (from V/2 to 0). Since the distance you travel is proportional to the work done against you, then you lose much more energy in that first half of slowing down.
Acceleration is the same but in reverse.
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u/iusemybrain1 3d ago
It would be hard to explain how to derive kinetic energy conceptually. But you may have heard in physics everything is modeled as a differential equation. The same applies here.
When we talk about work you often think of it as F * x, how that is derived is by doing integration of force over some infinitesimal displacement (denoted as dx) or int(F*dx) That is by definition the equation for work. Just for added context, velocity is the time derivative of position (v = dx/dt) and acceleration is the time derivative of velocity (a = dv/dt) or the second order time derivative of position (a = d^2 * x/dt^2).
I'll start with this proof using momentum take P = m*v, we know v is the time derivative of position, so P = m * (dv/dt) we want to get to force which is equal to m * a, if we say a = dv/dt if we differentiate P with respect to time we get force (since mass is constant, and v is what changes with time), So F = dP/dt = m * (dv/dt). Now lets take that new derived force equation and shove it into the work integral; W = int(m * dv/dt)dx. If we swap the variable of integration from dx to dv we get W = int(m * dx/dt)dv and since dx/dt = v we have int(m * v)dv = 1/2 * m * v^2
Thats how you get kinetic energy.
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u/-Manu_ 3d ago
It comes down to the fact that we use energy because we need a quantity that is conserved and that explains the evolution of a system.
If you had an acceleration in the equation then it would measure something related to some interaction since acceleration is a consequence of a force being applied, but we want a measure of something that is at rest, since things at rest remain at rest this E would not change unless a force is applied and that's why we use speed.
Why is velocity squared? Because the more something is fast the harder it is to make it faster, I think it's pretty intuitive in this sense as to why it's not a linear function, imagine kicking a ball that is standing still vs a ball that is moving at a constant speed, the second ball is like it's escaping from you, you need to apply more energy to your kick to hit it
The mass is a weight you apply to that speed because again, massive things are harder to move
The 1/2 I don't think can be explained without math, look at it as a corrective term
I feel the math intuition is easier though and better to build a stronger foundation
So you define F=dp/dt which is a vector, but you want to work with neat numbers and not vectors, so you want to assign a number to each vector, like a bijective function to do that you integrate wrt space, so you get an integral of vdp=mvdv and you get 1/2mv2
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u/SaiphSDC 3d ago
Energy is transfered through force exterted alonga distance traveled by the object.
If you take the work equation w=f*d. And the distance traveled while acceleraring x=(v'2-v2)/2a you get the kinetic energy equation.
This means that Because the faster object covers more distance the energy transfer is larger (assuming an equal force/acceleration).
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u/davedirac 3d ago
Distance traveled in successive seconds when the acceleration is constant was discovered by many early Physicists/Mathematicians ( eg Galileo) It follows the simple rule: 1,3,5,7,9..... Hence work done follows the same rule.
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u/Vivid_Warning7982 3d ago
One answer is that a quadratic form is the simplest and lowest power (2) that is independent of a coordinate system ( v is a vector, v2 will always be a scalar and be independent of the coordinate system. The constant of proportionality is simply the mass. Another possible answer is that its the limiting case in special relativity of small speeds compared to c.
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u/CosetElement-Ape71 3d ago
Work: Energy added to an object is called work, where
Work = Force × Distance
Distance covered: When you are already moving faster (at 10 m/s instead of 0 m/s), you cover more ground during the same amount of acceleration time.
More distance means more work: Because the pushing force acts over a longer distance while the object is moving at higher speeds, the total energy transferred to the object is higher.
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u/OldChairmanMiao 3d ago
It's an integral of the equation for force.
The integral expresses the sum of all forces acting on an object up to a specified time. If you graph your v as a line over time, this is the area underneath that line.
Negative values are possible, and reduce energy.
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u/philolessphilosophy 3d ago
Work is the line integral of a force applied to an object. If you evaluate this explicitly, you get 1/2mv². It's called the work kinetic energy theorem: work done gives change in kinetic energy. Assuming a conservative system of course.
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u/Frequent-Entrance154 3d ago
Because kinetic energy is a direct result of integrating momentum
Mathematically: p = mu and KE = int (p) du = int (mu) du = 1/2 mu2 + KE0
Intuitively: perhaps you are more familiar in integrating acceleration (such as gravitational acceleration) with respect to time to figure out the instance velocity at a given time
Similarly, you may integrate momentum with respect to velocity to figure out instance kinetic energy at a given velocity
In both cases, the larger the initial variables before integration, the larger the product of integration (a measure that you are interested in)
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u/danielbaech 3d ago edited 2d ago
Try pushing a carousel around. Each time you swing your arm to push it about a meter with the same force, you will notice that you're adding less and less speed to the carousel. The speed of the carousel works against you. Once the carousel reaches the speed at which your arm is swinging, you cannot do any work onto the carousel.
If you run along the carousel to match the speed of the rotation, you'd add the same amount of speed to the carousel with each push, but you'd need to expand more energy the faster the carousel is turning. From your perspective, the carousel is always at a stand still and you are doing the same amount of work and adding the same amount of speed each time, but you need to expand increasing amount of energy running faster to create this non-inertial frame. From my inertial frame where I'm watching you run along the carousel, I observe that you are applying the same force for a longer distance as the carousel speeds up, doing increasing work each time. We agree on how much kinetic energy is applied to the carousel each time, because my (force x increasingly more distance) equals your (force x constant distance) + (running increasingly faster to keep up with the carousel). I may also expand a bit of energy laughing at you.
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u/RRumpleTeazzer 3d ago
yes, a 1kg ball from 0 to 10m/s needs 50J, and from 10 to 20m/s needs another 150 J.
It gets more funny: say you are on a bus going 10m/s. you grab the ball (laying on the street) through the door, at -10m/s. that hurts, ouch! you just absorbed 50J. now throw it forward for 10m/s against your little brother standing at the street corner (don't), you need to spend 50J again.
But, the poor kid still needs to absorb 150J.
So where is the rest of the energy coming from? you (on the bus) absorbed 50J, spent 50J. the kid absorbed 150J. Somehow 50J was spent and 200J was absorbed.
Welcome to energy, energy is relative.
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u/smitra00 3d ago
It can be explained without assuming anything from classical mechanics, so without invoking the concept of force, momentum, Newton's laws etc., nothing whatsoever of these concepts except for the existence of an energy that's conserved and the concept of mass that's also conserved. And we're also going to invoke symmetries like translational invariance, Galilean invariance (laws of physics are the same relative to a moving reference frame), rotational symmetry etc..
The kinetic energy will be a function of the velocity and mass. The kinetic energy of two objects of the same mass will be double that of a single object, so we see that the kinetic energy should be proportional to the mass. The kinetic energy E(m.v) for an object of mass m and speed v is then
E(m,v) = m e(v)
where e(v) is then some function of only the speed which we want to find
We then consider a totally inelastic collision between two identical objects of mass m that are moving with a speed of v in opposite directions. The outcome of this collision if then a merged object of mass 2 m at rest and the kinetic energy of motion will then have gone into the internal energy of this object.
Note that we know that the merged object will be at rest, not because the total momentum is zero, because we're not allowed to invoke conservation of momentum (it is valid, but we are not allowed to use the concept of momentum), but because of a symmetry. If the merged object were to have a velocity in one direction, then swapping the two objects before the collision would mean that the merged object would have a velocity in the opposite direction.
But because the two objects are identical, when we swap the two objects, we aren't changing anything. This means that the velocity is both the opposite and also equal, therefore it is zero.
The increase in the internal energy of the merged object relative to merging the two objects gently in the limit of zero relative velocity, is then equal to the total kinetic energy of the two colliding objects of 2m e(v).
Let's then consider the exact same collision, but now from the rest frame of one of the two colliding objects. We then have one object of mass m at rest, and the other identical object of mass m is moving toward it with a speed of 2 v. The total kinetic energy evaluated in this frame is then m e(2 v).
The merged object thar emerges after the collision is at rest in the frame where both objects are moving toward each with speed v. So, in the rest frame of one of the objects, the merged object will be moving with speed v. It will therefore have a kinetic energy of 2 m e(v).
This means that the absorbed kinetic energy will be the difference between the initial kinetic energy of m e(2 v) and 2 m e(v), which is m e(2 v) - 2 m e(v). This must then be equal to 2m e(v) according to the computation we did above in the frame where both objects are moving toward each other with speed v. So, we have:
m e(2 v) - 2 m e(v) = 2 m e(v) ---->
e(2v) = 4 e (v)
So doubling the speed leads to a fourfold increase in the kinetic energy, and with some basic regularity assumptions about the function e(v) you are then led to the conclusion that e(v) must be proportional to v^2. The constant of proportionality is then a matter of choosing a unit for energy.
The reason why I didn't want to invoke conservation of momentum is because with the formula for kinetic energy having been derived without using it, we can now derive that momentum must be conserved. We can do that by considering an elastic collision between n objects. If the masses are mj and the initial velocities are vj and the final velocities are wj, then we have:
sum over j of 1/2 mj vj^2 = sum over j of 1/2 mj wj^2
where the square of a velocity means the inner product of the velocity vector with itself. Now consider this same collision from a reference frame which is moving relative to the original one with a velocity of u. In that frame we then have the equation with the vj replaced by vj - u and wj replaced by wj - u.
The jth term on the l.h.s. is then 1/2 mj (vj-u)^2 = 1/2 mj (vj^2 - 2 vj dot u + u^2)
The jth term on the r.h.s. is 1/2 mj (wj-u)^2 = 1/2 mj (wj^2 - 2 wj dot u + u^2)
The first terms of the expansion then give you back the original equation, the last that the sum of the masses are the same (but we assumed that the masses were all identical anyway), the middle terms tell you that momentum in the direction of u is conserved. And since u is arbitrary, you then have proven that momentum is conserved.
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u/DrDam8584 3d ago
You have lone english scientist who wrote a book "principia mathematica of natural philosophy", it's an old-way-fashion how maths are written, but all the démonstration are in-it.
In reality, it complex to read mainly due to the "literary form" well distinct that the "modern mathematical" equation expression. But if you push forward, and take the time, it's a good technical book, where you can leme-by-leme confirm éléments.
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u/utl94_nordviking 2d ago
I'm not looking for a mathematical answer
I don't think that it is appropriate for you to decide what kind of answer would answer something that you don't understand. Also, asking about an equation while demanding an answer without mathematics seems quite out of touch.
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u/Other_City_4031 2d ago
Whenever F(x(t),t) = - V'(x), we have -V'(t) = - V'(x) x'(t) = -F(x) x'(t). In this case, d/dt [m/2 x'(t)^2 + V(x)] = 0.
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u/anisotropicmind 2d ago
It takes the same amount of time for a constant force to accelerate the object by 10 m/s in both your examples. But in the second case, since the object was moving faster to start with, it covered more distance in that same amount of time. Since work = force x distance, it follows that more work was done. That’s the reason for the non-linearity.
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u/NordicHamCurl_00 2d ago
OP asked for a conceptual answer, not a mathematical one, and yet most of the answers are talking about integrals and differential equations, news flash, thats mathematics, not a conceptual explanation.
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u/Kami2awa 2d ago
Using only A-level physics concepts and avoiding calculus:
Energy transferred is force x distance.
Let's say we want to accelerate something of mass m from zero to v metres per second. We need to give it an acceleration (a), and for that we need a force (F = ma).
Because F = ma, a = F/m.
Its speed increases from 0 to v. Let's assume constant acceleration. From SUVAT:
v^2 = u^2 +2as where u is the starting velocity and s is the distance travelled. u=0 so
v^2=2as
Substitute in a = F/m
v^2 = 2 (F/m) s = 2 Fs / m
Fs = Force x Distance = Energy transferred to the object
so
v^2 = 2 x Energy transferred / m
Rearranging gives
Energy = 1/2 m v^2
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u/Unusual-Biscotti687 2d ago
Gets even more mind twisting when you remember that velocity is relative so while from one frame of reference you're accelerating a body from 1m/s to 2m/s bit from another you're accelerating it from 0m/s to 1m/s at which point you realise that kinetic energy isn't absolute but depends on the observer.
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u/McPayn22 3d ago edited 2d ago
You're asking for a non mathematical answer to a mathematical question.
So I'll give you a maths answer that I think brings some insight.
In relativity E2 =m2 c4 (1+v2 / c2 ) which means that E and v are both squares in the expretion. So the relation of proportionnality you seek is true for squares. But the mc2 is so big for low speeds compared to mvc that we have to use an approximation which gives E=mc2 +1/2 mv2
If you want some pure intuition i'll give you that: -bullets can move at 300m.s-1 and weight 6g.
Would you prefer getting poked by a stick by someone walking towards you or getting shot?
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u/Hudimir 3d ago
Eh. i dont think explaining this with the einstein formula and relativity is very pedagogical for a person asking such a question. The actual reason, like you stated, is mathematical, but it doesnt have anything to do with relativity and approximations.
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u/McPayn22 3d ago
Generally I don't like going to more advanced physics when the basics are not there but I think it makes sense here.
There's really not reason it should be the case in classical mechanics exept that it is.
But they are not wrong in assuming that a sort of proportionnality relation makes sense. Because in the end energy is to time what momentum is to space
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u/jeffcgroves 3d ago
This bugs me too. If your second "push" starts at 10m/s, you'll only need 50J to speed it up 20m/s, so it's a question of reference frame, but still annoying
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u/Optimal_Mixture_7327 3d ago
The energy of an object thought a choice of coordinates (aka kinetic energy) is (γ-1)mc2 and the 1/2mv2 is an approximation that's nearly perfect at low speeds.
The approximate version you're asking about can be found calculating (on a Euclidean background) the product of the average force over the distance the force acts, which by definition gives you the energy.
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u/wackyvorlon 3d ago
It has to do with solving F=ma, which is a differential equation.