How Do You Graph Kinetic Energy Against Time and Distance?

Click For Summary
SUMMARY

The discussion focuses on calculating and graphing kinetic energy (KE) for a mass of 6 kg subjected to a horizontal force of 80 N over a distance of 5 m in 0.92 seconds. The total energy expended during acceleration is calculated to be 400 J, with the kinetic energy at the end of the interval determined to be approximately 354.44 J. The graphs of kinetic energy against time and distance are discussed, with the expectation that the KE vs. time graph will exhibit a parabolic shape, while the KE vs. distance graph will be linear. The user clarifies their understanding of the relationship between the two graphs and seeks confirmation on their graphical representations.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with kinetic energy formula: KE = 1/2 mv²
  • Basic knowledge of graphing techniques in physics
  • Ability to perform calculations involving acceleration and force
NEXT STEPS
  • Learn how to graph kinetic energy as a function of time using Excel or graphing software
  • Study the relationship between force, mass, and acceleration in detail
  • Explore the concept of work-energy theorem in physics
  • Investigate how to interpret and analyze different types of graphs in physics
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of kinetic energy and graphing in physics.

james123
<Moderator's note: Moved from a technical forum and thus no template.>

A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5 m in 0.92 s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant:
  1. (a) Calculate the total energy expended in the acceleration.

    (b) Plot a graph of the kinetic energy of the mass against time.

    (c) Plot a graph of the kinetic energy of the mass against distance. Work done so far:

    1. (a) 80 x 5= 400J

    1 (b) a= 2d/t^2
    a=(2x5)/0.92^2
    a=11.8147448
    a=11.81 m/s^-2

    v=at
    v=11.8147448 x 0.932
    v=10.86956522

    KE=1/2m(at)2
    KE=0.5 x 6(10.86956522)^2
    KE=354.4423442 J

    1 (c) KE=1/2m(2ad)
    KE=0.5 x 6(2 x 11.8147448 x 5
    KE=354.442344 JMy main questions are, should I expect the same answers for both questions? seems a little odd to me.
    Also, I'm completely stumped on the graphs, how to start them, how to plot them etc.

    Any help is greatly appreciated!
 
Last edited by a moderator:
Physics news on Phys.org
james123 said:
should I expect the same answers for both questions? seems a little odd to me.
Why ? There is only one kinetic energy at t = 0.92 s, which is what you calculated in b) and c). What the exercise asks is a plot of KE vs t in b) and vs d in c). So far you have two points (t=0 and t = 0.92 s) for the first and two points (d=0 and d=5 m) for part c).

Please post in the homework section and use the template

Oh, and if the exercise given data are in two digits, don't report in 10 digits (calculating in that many digits is just fine though, with todays equipment :smile:).
 
Hi, thanks for replying.

I have now posted in the homework section.

I think I was just suspicious about getting two of the same answers, I assumed I was doing something wrong.

After thinking about it, it now makes sense, just got to figure out these graphs!
 
My first graph for 1(b) has ended up looking like this, is this along the right lines??
untitled-jpg.jpg
 
Does it look like a graph of
james123 said:
KE=1/2m(at)2 ?
And: does it show
james123 said:
KE=354 J
at t = 0.92 s ?
 
Sorry, the graph I posted was generic and isn't mine, just used it as an example to show the general shape of what I ended up with.

That said, I've re-done the graphs and think I have it. Please ignore the values as again these aren't my graphs.

For question 1 (b) I have this graph where KE grows exponentially.
upload_2017-8-10_13-58-44.jpeg

And for 1(c) I have needed up with a straight line graph

untitled-jpg-jpg.jpg


Still trying to figure out how to upload my own pictures from a MacBook! Are these the shaped of graph I should expect??

Thanks for chipping in so far, I really appreciate it
 
james123 said:
Are these the shaped of graph I should expect?
Yes. a parabola for KE=1/2m(at)2 and a straight line for KE=1/2m(at)2
 

Similar threads

Replies
13
Views
18K
Replies
38
Views
4K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
22
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 8 ·
Replies
8
Views
2K