# How is acceleration = work done?

Hi

So, let's go straight to the point: a = v - u / t
Kinetic Energy: 1/2 mv2

Ok, I understand that a acceleration, a change in velocity, will change the Kinetic Energy since v is part of the Kinetic energy equation.

But acceleration is not simply a change in velocity. It is the rate, time is involved.

So, I don't understand how we can use Kinetic Energy to answer a question like this for example: Assuming that all the power of the motor is used to accelerate the cars, calculate how long they will take to reach their maximum speed.

For the question above though, we are given the power, is it correct to say therefore that:

Acceleration = Rate of change in Velocity, a change in velocity, is equal to a change in Kinetic Energy, which is equal to work done.

Thus, Power being: Work Done / t

We have the work done, which is equal to Kinetic Energy, a change in velocity but also, over time, which is acceleration.

I hope I made my doubt clear. If not, please tell me so I can rephrase.

Related Introductory Physics Homework Help News on Phys.org
there is a flaw in your argument. an acceleration will always produce a change in velocity, but will not always produce a change in speed. if the acceleration is perpendicular to the velocity, as in circular motion, then only the direction of the velocity will change, not the magnitude. thus, the kinetic energy does not change and so no work is done.

hope this helps clear things up.

cheers

Hi eczeno, thanks for your response

Ok, so, provided the acceleration is not at 90 degrees to the velocity, an acceleration will cause a change in velocity, which is equal to a change in kinetic energy, hence, work done.

But, acceleration being the rate of change in velocity is equal to power:

Power = Work Done / time

Considering Work Done to be a change in velocity.

Is that ok?

As you said, power is work done over time. In the case of a moving object like a car, work is related to the force acting on the car and the distance the car travels. The force acting on the car is related to the mass of the car, and the acceleration of the car.

Now, I think your question is how is change in kinetic energy related to acceleration?

You're right, change in velocity on its own does not define acceleration. However, it turns out that when you have a change in the squared velocity, this becomes something similar to a change in displacement (not over time) and a change in velocity (over time). See the derivation in this post: https://www.physicsforums.com/showpost.php?p=3270748&postcount=3. Near the end of that derivation, the change in kinetic energy is expressed as a function of acceleration. Hopefully that helps.

I understood the derivation. It was helpful to understand how a Change in Kinetic Energy is related to work.

But, sorry, I am still a bit confused on how it Kinetic Energy is related to acceleration.

hi pro,

i suggest you take a look at greenlaser's post, it is a nice derivation. one thing to remember is that work and acceleration are defined quantities, and they are defined differently. in particular, one is a vector and one is a scalar, so it does not make mathematical sense to equate them.

cheers

one way to think about it is that we are approaching the problem from two directions. Vector dynamics (force, acceleration, velocity), and energy, which is a scalar approach. these approaches must be connected, because they describe the same physical system, but the connections can be hidden behind some complicated math. It is unfortunate that these concepts are introduced in a way that does not always make those connections clear, but that is necessary to reduce the mathematical rigor. The link above is a small step in the direction of mathematical rigor.

it turns out that kinetic energy is related to the speed of the object, but speed is only part of velocity, there is also the direction, which kinetic energy simply does not care about. acceleration, on the other hand, cares about both parts of the velocity. thus, acceleration can tell us a lot about the kinetic energy, but if we try to get the acceleration from the change in KE, we cannot retrieve any information about the direction. what i have just said simplifies the issue too much, it's not the whole picture, but it should help bring it into focus a little.

cheers

olivermsun
I understood the derivation. It was helpful to understand how a Change in Kinetic Energy is related to work.

But, sorry, I am still a bit confused on how it Kinetic Energy is related to acceleration.
Suppose you toss a rock upward. It started with some amount of KE, but gravity acts against the velocity and slows it down. Hence KE = 1/2 mv^2 is actually decreasing (as it turns out, the KE is being traded for potential energy).

At some point the rock will turn around and start falling downward. Now the acceleration is in the same direction as the velocity, and it begins gaining KE again.

The point is, there must be some acceleration in the same direction that something is already going to make it go faster and increase KE.

Hi all,

Yes, eczeno your simplification as you called it did help

I recognize I am just generally confused and the relations might not be so clearer.

But just to make things a little bit clearer, and it is the last thing I am going to ask

I am going to use that question again in order to explain my thought: Assuming that all the power of the motor is used to accelerate the cars, calculate how long they will take to reach their maximum speed.

So the acceleration bit is essentially: The work done, the change in kinetic energy for the "magnitude" part of the change in velocity while the Power part kicks in to be the rate part: The work done would be the velocity change part in acceleration while the time part of the Power would be the rate?

The change in kinetic energy does in fact contain acceleration and you do not need the power part to make it into a rate. I'm going to go back to the algebra I linked to, but I'm going to focus more on what you're asking, hopefully making it clearer.

$$K_{net} = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}m (v_f^2 - v_i^2) = \frac{1}{2}m(v_f - v_i)(v_f + v_i) = \frac{(v_f + v_i)}{2} (v_f - v_i) m$$
All that is just algebraic manipulation of the change in kinetic energy; nothing mysterious there. However, note the next step. Thanks to the "squaredness" of the velocity, you have two terms with velocity. You've got the $$(v_f - v_i)$$ term, which is the change in velocity you're talking about, but you also have an average velocity term:
$$\frac{(v_f+v_i)}{2}$$

Average velocity is a rate of change with respect to time. When you turn it into this rate form,
$$\frac{(v_f+v_i)}{2}= v_{avg} = \frac{\Delta d}{\Delta t}$$

You get your missing "time change" term.

So,
$$K_{net} = (v_{avg})(v_f - v_i) m = \frac{\Delta d}{\Delta t}(v_f - v_i)m = m\Delta d \frac{(v_f - v_i)}{\Delta t}$$

Now you've got a change in velocity over a change in time: a rate, which does define acceleration, unlike the $$(v_f - v_i)$$ term alone (as you said). So now we can say,
$$K_{net} = m\Delta d a$$

Hopefully that clears up how you can have acceleration based on change in kinetic energy. You're correct in saying a difference in velocity does not represent acceleration, because you need to know how fast that change takes place. However, since you have a change in velocity squared you have two rates with time; two terms with "change in position over change in time" (speed) terms. You can convert those to have one term with a change in position over a change in time, over a change in time (acceleration) through algebraic manipulation.

Unfortunately, as eczeno mentioned, the math is a little bit tricky because we're relating scalars and vectors. However, if we assume that everything is acting in one line, or is one-dimensional, then everything works as a scalar, and the math actually works out quite nicely without having to use calculus and vector math. You always have to be very careful what assumptions you're making, because kinetic energy isn't always directly related to an acceleration, it just works out in the one dimensional case. Don't take that to mean all of this is meaningless though, because understanding simple cases and simple relationships is crucial to understanding complex relationships.

Nice! Thanks!

No problem.