# Kinetic energy and work confusion [closed]

I know that $\frac{1}{2}mv^2$ is derived from $W = \int \vec F \cdot\ d\vec s$, but are there exceptions to the kinetic energy formula? Newton says that unless an external force is applied, a body will remain at rest or in uniform motion (first law of motion). This means that its velocity will be constant, but now problems start.

A car of $2\,\mathrm{kg}$ has a constant velocity of $1\,\mathrm{m/s}$ (in a frictionless planet). At point A, it is $1\,\mathrm{m/s}$ and at point B (after $10\,\mathrm{m}$) the velocity is still $1\,\mathrm{m/s}$. We say that the change in kinetic energy is the one with $V_f - V_i$. Therefore the net $K.E.=W=0\,\mathrm{J}$ because both the intial and final velocity are the same, and so $1\,\mathrm{J}-1\,\mathrm{J}=0\,\mathrm{J}$. This seems ok because I'm not applying any force (its frictionless, remember). SO, $W = 0\,\mathrm{N}(10\,\mathrm{m})=0\,\mathrm{J}$.

But, what if it is in a friction planet and in order to overcome the force of friction is $4 N$, then our equations contradict. $W=(4\,\mathrm{N})(10\,\mathrm{m})= 40\,\mathrm{J}$. Why do I get answers that contradict each other?

(2nd question) If a car of $4\,\mathrm{Kg}$ travels in New York in a road (During $10\,\mathrm{m}$, with friction of course) at a constant velocity of $2\,\mathrm{m/s}$, we say it has kinetic energy of $8\,\mathrm{J}$, because $\frac{1}{2} (4)(2)^2=8\,\mathrm{J}$. What if the force applied to overcome the force of friction is just $2\,\mathrm{N}$, then we say that during those $10 m$ its $$\text{energy}=Fd=(2\,\mathrm{N})(10\,\mathrm{m})=20\,\mathrm{J}$$Again, I got different answers.

My last question is, If you don't apply force, you don't have energy, right? So, in a frictionless place, if an object is moving, is it MOVING because it has energy or not?

## closed as unclear what you're asking by ACuriousMind♦, DanielSank, Kyle Kanos, John Duffield, Bill NOct 5 '15 at 20:44

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Man, there are a lot of misconceptions and problems contained in your post. I'm not sure where to start. Also, a lot of the thoughts in your post are not clearly expressed. For one, your statement of Newton's First Law of Motion is rather garbled. Why don't you try to organize your thoughts better and distill them down to a key question or two and then go from there? – Samuel Weir Oct 4 '15 at 1:18
• Jeps, please check your question. I have edited certain parts. It still requires editing and only you can do it considering what exactly you want to ask. – SchrodingersCat Oct 4 '15 at 12:44
• Welcome to Physics Stack Exchange. On this site, please ask one specific question per post. Asking multiple questions at once makes it considerably more difficult to write an appropriate answer, and means that you won't get the help you're looking for. – DanielSank Oct 4 '15 at 19:29
• You have assumed that $KE = W$. That's not true. What you should have is $\Delta KE = W$. – Bill N Oct 5 '15 at 20:43

You get different answers because these are different situations.

let's start with the first case. In the frictionless planet, speed is constant if the sum of all forces acting over the body are 0. No difference in kinetic energy, no work done.

In the second case, the work should be negative (the friction force opposes the displacement, work is the dot product of force and displacement). This makes sense, since the body will have deaccelerated (because there is a friction force). Then final speed squared is lower than initial speed squared, so work is negative. Also, depending on the body's mass and initial speed, it could be that it stops before reaching B.

If you want to keep the body moving with constant speed while it is subject to a friction force, then you have to apply a force that counters the friction. The work done by this force is positive and equal (in absolute value) to the work done by the friction force, keeping total work at 0 (and no change in the speed, so again, work is 0).

In the third case, the car is travelling with constant speed, so the forces are balanced. As I explained before, the total work is 0, because the forces have equal absolute value but are acting in opposite directions, over the same displacement.

But, what if it is in a friction planet and in order to overcome the force of friction is 4N, then our equations contradict. W=(4N)(10m)=40J. Why do I get answers that contradict each other?

If it's on a surface with friction, then the final and start velocity will no longer be the same. You calculated $40\,\mathrm{J}$, so the kinetic energy must decrease with $40\,\mathrm{J}$, which gives a lower speed.

If a car of 4Kg travels in New York in a road (During 10m, with friction of course)¨

Remember that a car is kind of a special case here. There is namely no friction (ideally) when a car moves. This is because the wheels turn, which is a different case than if a box simply slided over the surface.

If not ideal cases, then yes, there is some ("rolling") friction.

What if the force applied to overcome the force of friction is just 2N, then we say that during those 10m its energy=Fd=(2N)(10m)=20J Again, I got different answers.

Why did you choose $2\,\mathrm{N}$ here? If friction does $-40\,\mathrm{J}$ of work on the car, then the engine must add the same amount $40\,\mathrm{J}$ to overcome this to maintain the kinetic energy and avoid a decrese. If the engine does not do that, then the car will slowly stop as kinetic energy gets lower and lower, no doubt about that.

My last question is, If you don't apply force, you don't have energy, right?

No, this in not true. A space ship that is simply flying in outerspace (no forces to stop it) can turn off it's engines and will continue to move with constant speed (as you explained in your example, speed is constant when no forces are applied). It continues to have it's kinetic energy.

Energy and force are not both required at the same time. You can have a table applying a force to hold up a book, without spending any energy. And you can have kinetic energy without a force as explained.

You need forces to change the energy states (like slowing down), but energy being present does not require a force to be present also.

So, in a frictionless place, if an object is moving, is it MOVING because it has energy or not?

Rather say it like this: If it moves, then it does have (kinetic) energy. Kinetic energy and movement are by definition two sides of the same coin.