Why Is the Derivation of Kinetic Energy Confusing?

In summary, the conversation discusses the derivation of kinetic energy from work and raises two questions related to the manipulation of equations and the calculation of the change in kinetic energy between two points. The first question is about changing the division between (2) and (3), while the second question is about the correct representation of the change in kinetic energy. The conversation concludes with an explanation of the validity of the transposition in question (1) and the incorrectness of the result in step (4).
  • #1
bentley4
66
0
Hi everyone,

There are 2 things I do not understand in the derivation of kinetic energy from work:
https://www.physicsforums.com/showthread.php?t=111162

(1) W = [itex]\int[/itex][itex]\vec{F}[/itex](t).d[itex]\vec{r}[/itex](t)=

(2) m.[itex]\int[/itex][itex]\frac{d\vec{v}(t)}{dt}[/itex].d[itex]\vec{r}[/itex](t)=

(3) m.[itex]\int[/itex]d[itex]\vec{v}[/itex](t).[itex]\frac{d\vec{r}(t)}{dt}[/itex]=

(4) [itex]\frac{m}{2}[/itex].(v(t1) - v(t0))[itex]^{2}[/itex]

Question I: I don't understand why you can just change the division like that between (2) and (3). I know multiplication and division have the same order but they are calculated from left to right. So why can you do that from (2) to (3)?
http://en.wikipedia.org/wiki/Order_of_operations

Question II: I don't understand why the change in kinetic energy between t1 and t0 is sometimes written as [itex]\frac{1}{2}[/itex].m.v(t1)[itex]^{2}[/itex] - [itex]\frac{1}{2}[/itex].m.v(t0)[itex]^{2}[/itex]
E.g. like in https://www.physicsforums.com/showthread.php?t=194461.

After (4), it should be [itex]\frac{m}{2}[/itex].v(t1)[itex]^{2}[/itex] - m.v(t1).v(t0) + [itex]\frac{m}{2}[/itex].v(t0)[itex]^{2}[/itex] , right?
 
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  • #2
bentley4 said:
Question II: I don't understand why the change in kinetic energy between t1 and t0 is sometimes written as [itex]\frac{1}{2}[/itex].m.v(t1)[itex]^{2}[/itex] - [itex]\frac{1}{2}[/itex].m.v(t0)[itex]^{2}[/itex]
E.g. like in https://www.physicsforums.com/showthread.php?t=194461.

After (4), it should be [itex]\frac{m}{2}[/itex].v(t1)[itex]^{2}[/itex] - m.v(t1).v(t0) + [itex]\frac{m}{2}[/itex].v(t0)[itex]^{2}[/itex] , right?
No. Your step (4) is incorrect.
 
  • #3
Doc Al said:
No. Your step (4) is incorrect.

I'm sorry, I don"t understand what I am doing wrong going from (3) to (4). Can you elaborate please?
 
  • #4
bentley4 said:
I'm sorry, I don"t understand what I am doing wrong going from (3) to (4). Can you elaborate please?
∫v dv = v2/2 → v12/2 - v02/2
 
  • #5
Re your QI...
I respect your unwillingness to regard [itex]\frac{d\vec{v}}{dt}[/itex] as one thing divided by another, but you might be happier to do this with [itex]\frac{Δ\vec{v}}{Δt}[/itex], and then associate Δt with Δ[itex]\vec{r}[/itex] to make [itex]\vec{v}[/itex]. Only then take it to the limit.

I don't suppose this is rigorously valid, but I suspect this is how a lot of physicists justify this sort of transposition.
 
  • #6
Thnx for your responses.
 
  • #7
bentley4 said:
Hi everyone,

There are 2 things I do not understand in the derivation of kinetic energy from work:
https://www.physicsforums.com/showthread.php?t=111162

(1) W = [itex]\int[/itex][itex]\vec{F}[/itex](t).d[itex]\vec{r}[/itex](t)=

(2) m.[itex]\int[/itex][itex]\frac{d\vec{v}(t)}{dt}[/itex].d[itex]\vec{r}[/itex](t)=

(3) m.[itex]\int[/itex]d[itex]\vec{v}[/itex](t).[itex]\frac{d\vec{r}(t)}{dt}[/itex]=

(4) [itex]\frac{m}{2}[/itex].(v(t1) - v(t0))[itex]^{2}[/itex]

Question I: I don't understand why you can just change the division like that between (2) and (3). I know multiplication and division have the same order but they are calculated from left to right. So why can you do that from (2) to (3)?
http://en.wikipedia.org/wiki/Order_of_operations

Question II: I don't understand why the change in kinetic energy between t1 and t0 is sometimes written as [itex]\frac{1}{2}[/itex].m.v(t1)[itex]^{2}[/itex] - [itex]\frac{1}{2}[/itex].m.v(t0)[itex]^{2}[/itex]
E.g. like in https://www.physicsforums.com/showthread.php?t=194461.

After (4), it should be [itex]\frac{m}{2}[/itex].v(t1)[itex]^{2}[/itex] - m.v(t1).v(t0) + [itex]\frac{m}{2}[/itex].v(t0)[itex]^{2}[/itex] , right?

Question I: Because a derivative is equivalent to a division of differentials. And then by usual algebraic rules you can move the denominator.

Question II: It is not sometimes, but always, because that is the result of the integration of (3). You own result (4) is incorrect.
 
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1. What is the definition of kinetic energy?

Kinetic energy is the energy an object possesses due to its motion.

2. What is the equation for calculating kinetic energy?

The equation for kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. How is kinetic energy derived from Newton's laws of motion?

According to Newton's second law of motion, the force acting on an object is equal to its mass multiplied by its acceleration. Using this, we can derive the equation for kinetic energy by integrating the equation for force over distance.

4. What are the units of kinetic energy?

The units of kinetic energy are joules (J) in the SI system and foot-pounds (ft-lb) in the imperial system.

5. How is kinetic energy related to potential energy?

Kinetic energy and potential energy are two forms of mechanical energy. Kinetic energy is associated with an object's motion, while potential energy is associated with its position in a force field. The total mechanical energy of an object is the sum of its kinetic and potential energy.

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