# Solving a V-vs-s Problem: John's Quandry

• John Mohr
In summary, John's error was in writing the work-energy theorem incorrectly. He has written the equation incorrectly, and the wrong side will only be equal to the change in kinetic energy if v_i=0 or v_f=v_i.
John Mohr
Homework Statement
A 2500 kg car changes from a velocity of 15 m/s to 25 m/s over a distance of 450 m. What is the work done on the car by the engine?
Relevant Equations
W = Fd, Delta E = 1/2mdeltav^2
W = delta E
Hello,
For some reason I'm not seeing something in working out a question two ways. The question I'm working through reads:

"A 2500 kg car changes from a velocity of 15 m/s to 25 m/s over a distance of 450 m. What is the work done on the car by the engine?"

I've attached my quandry. I've surmised that the work-energy theorm can be used. Furthermore, that on the work side, I could use kinematics and on the energy side, find the change in kinetic energy. But I don't come up with the same answer on each side? This is mostly because of the difference between (vf^2-vi^2) and (vf-vi)^2. I've attached some handwritten work.

Does anyone see my error?

John

#### Attachments

• Kinematics vs Energy.pdf
120.3 KB · Views: 76
In your handwritten work, you did not calculate the change in kinetic energy correctly.
## KE_f - KE_i \neq \frac 1 2 m (\Delta v)^2##.

John Mohr
Very good! For some reason, I've used 1/2m(deltav)^2 in the past. This is my mistake.
Is there any use for this expression or is this completely wrong and never used?

John Mohr said:
Homework Statement:: A 2500 kg car changes from a velocity of 15 m/s to 25 m/s over a distance of 450 m. What is the work done on the car by the engine?
Relevant Equations:: W = Fd, Delta E = 1/2mdeltav^2
W = delta E

Hello,
For some reason I'm not seeing something in working out a question two ways. The question I'm working through reads:

"A 2500 kg car changes from a velocity of 15 m/s to 25 m/s over a distance of 450 m. What is the work done on the car by the engine?"

I've attached my quandry. I've surmised that the work-energy theorm can be used. Furthermore, that on the work side, I could use kinematics and on the energy side, find the change in kinetic energy. But I don't come up with the same answer on each side? This is mostly because of the difference between (vf^2-vi^2) and (vf-vi)^2. I've attached some handwritten work.

Does anyone see my error?

John
Your error is that you wrote the work-energy theorem incorrectly. You have

The right hand side should be the change in kinetic energy ##\Delta K= \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2##, i.e. subtract the initial KE from the final KE.

John Mohr
John Mohr said:
Very good! For some reason, I've used 1/2m(deltav)^2 in the past. This is my mistake.
Is there any use for this expression or is this completely wrong and never used?
I don't know of any use for the expression ##\frac 1 2 m (\Delta v)^2##. I recommend that you trash it.

malawi_glenn and John Mohr
TSny said:
I don't know of any use for the expression ##\frac 1 2 m (\Delta v)^2##.
Thank you so much for your help!

TSny
@TSny @kuruman
This is crazy to realize for me. I've been teaching it this way for years and many other teachers I know teach it this way too. This video segment is a prime example. It's amazing how these things make their way into our misunderstandings.
Thanks again.
John

hutchphd, PeroK and TSny
John Mohr said:
@TSny @kuruman
This is crazy to realize for me. I've been teaching it this way for years and many other teachers I know teach it this way too. This video segment is a prime example. It's amazing how these things make their way into our misunderstandings.
Thanks again.
John
Yes, the video is incorrect in the way it solved for ##\Delta v##. The final answer for ##v_f## happened to come out right because ##v_i = 0## in the example.

##\Delta (v^2)## will equal ##(\Delta v)^2## only if ##v_i = 0## or if ##v_f = v_i## (trivial).

Note that ##\frac 1 2 m (\Delta v)^2## is always nonnegative. But we know we can have situations where the KE decreases; that is, ##\Delta KE## is negative.

John Mohr
John Mohr said:
@TSny @kuruman
This is crazy to realize for me. I've been teaching it this way for years and many other teachers I know teach it this way too. This video segment is a prime example. It's amazing how these things make their way into our misunderstandings.
Thanks again.
John
I cannot make that link work. Just stuck on some ad that won't play.
Seems to me you confused ##\Delta(v^2)## with ##(\Delta(v))^2##.

Lnewqban and John Mohr
John Mohr said:
@TSny @kuruman
I've been teaching it this way for years and many other teachers I know teach it this way too.
This is scary, if true. What kind of teachers? Grade school teachers?

malawi_glenn
John Mohr said:
@TSny @kuruman
This is crazy to realize for me. I've been teaching it this way for years and many other teachers I know teach it this way too. This video segment is a prime example. It's amazing how these things make their way into our misunderstandings.
Thanks again.
John
On the segment you supply the instructor explicitly emphasizes that the form he wrote is good only for v(initial)=0. You should indeed be concerned.

haruspex said:
I cannot make that link work. Just stuck on some ad that won't play.
Seems to me you confused ##\Delta(v^2)## with ##(\Delta(v))^2##.
I got a "Skip Ad" 5 s count down timer on the video lower right.
haruspex said:
Seems to me you confused ##\Delta(v^2)## with ##(\Delta(v))^2##.
I find the interpretation of the term ##\Delta (v^2)## ambiguous. It is obvious from the context how you mean it here but, when doing error analysis, one also writes ##\Delta (v^2)=2v\Delta v##. That could be confusing to beginners who have not yet mastered the subtleties of when one uses just a "##-##" sign, a "##\Delta##", a "##\delta##" or a "##d##" to denote a difference between two quantities.

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