- #1
Miran97
- 1
- 0
Hi!
The kinetic formula are EK=1/2*m*v^2. But how do we find it?
The kinetic formula are EK=1/2*m*v^2. But how do we find it?
The Power of Kinetic Energy to the EK=1/2*m*v^2 Formula
- Thread starter Miran97
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AI Thread Summary
The discussion centers on the kinetic energy formula EK=1/2*m*v^2 and its derivation. The derivation involves understanding how work is done to accelerate an object using Newton's second law and the definition of work. The key steps include relating force, mass, and acceleration, ultimately leading to the conclusion that the work done to reach a certain velocity is equal to the kinetic energy. There is also a mention of the challenge in clarifying the actual questions or misunderstandings of those learning physics. Understanding the underlying concepts is essential for effective explanation.
- #2
Doc Al
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Looks like you already found it!
What do you mean? (How you would find the kinetic energy in a particular problem will depend on the details of that problem.)
What do you mean? (How you would find the kinetic energy in a particular problem will depend on the details of that problem.)
- #3
lep11
- 380
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Maybe he/she meant to ask how the formula is derived.
- #4
K^2
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The simplest derivation is to consider how much work is done to accelerate an object to velocity v. Let's say we use constant force to accelerate the object over some distance d.
F = ma -> a = F/m (From Newton's 2nd)
W = F*d (From definition of work.)
v² = 2*a*d (Let us know if you need help deriving this one, but you should have seen it.)
Putting it all together.
v² = 2*d*a = 2*d*(F/m) = 2*(d*F)/m = 2*W/m
W = (m*v²)/2
So the amount of work required to accelerate (or decelerate, for that matter) an object of mass m to velocity v is (1/2)*m*v². Hence the Kinetic Energy.
F = ma -> a = F/m (From Newton's 2nd)
W = F*d (From definition of work.)
v² = 2*a*d (Let us know if you need help deriving this one, but you should have seen it.)
Putting it all together.
v² = 2*d*a = 2*d*(F/m) = 2*(d*F)/m = 2*W/m
W = (m*v²)/2
So the amount of work required to accelerate (or decelerate, for that matter) an object of mass m to velocity v is (1/2)*m*v². Hence the Kinetic Energy.
- #5
Doc Al
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Anyone else want to guess what the OP's real question was? 
- #6
K^2
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I've always found that to be the real challenge of explaining physics to someone. Once you understand what it is that the person actually wants to know or doesn't understand, explaining it is easy.Doc Al said:Anyone else want to guess what the OP's real question was?
Hello everyone,
Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed.
My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds?
Best regards,
DaTario
Hello!
I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below.
Here is how I wrote the condition of equilibrium forces:
$$
\begin{cases}
F_{g\parallel}=F_{t1}+F_{t2}, \\
F_{g\perp}=F_{r1}+F_{r2}
\end{cases}.
$$
On the other hand...
I know that mass does not affect the acceleration in a simple pendulum undergoing SHM, but how does the mass on the spring that makes up the elastic pendulum affect its acceleration? Certainly, there must be a change due to the displacement from equilibrium caused by each differing mass? I am talking about finding the acceleration at a specific time on each trial with different masses and comparing them. How would they compare and why?
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