Work question without distance, work and velocity

  • Thread starter Thread starter MsHomework
  • Start date Start date
  • Tags Tags
    Velocity Work
Click For Summary
SUMMARY

The discussion focuses on calculating the work done on an 8.0 kg wagon that is brought to rest by a 60 N force over 2 seconds. The key equations involved include Newton's second law (F = ma) and the work-energy principle (W = F · d). Participants clarify the relationship between force, mass, and acceleration, and derive the initial velocity and distance traveled using kinematic equations. The final work done is confirmed to be 900 Joules, demonstrating the application of physics principles in problem-solving.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with kinematic equations (SUVAT equations)
  • Knowledge of work-energy principle (W = F · d)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation and application of the SUVAT equations in various motion scenarios
  • Explore the work-energy theorem and its implications in physics problems
  • Practice problems involving force, mass, and acceleration calculations
  • Learn about the concept of momentum and its relation to force and motion
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of work, force, and motion in real-world applications.

  • #31
MsHomework said:
velocity final^2= velocity initial^2 + 2(acceleration)(distance)

This one would work, but there is another one with which you can calculate the distance directly.
Can you find it?
 
Physics news on Phys.org
  • #32
d= 1/2 (velocity initial + velocity final)t
 
  • #33
MsHomework said:
d= 1/2 (velocity initial + velocity final)t

This is the one formula I can't make sense of.
I can't think of any context where you would want to use it.

Let's not use it.

Do you have yet another one?
 
  • #34
Ok well I decide to just do v=at. then I did velocity final= velocity initial + (acceleration)(distance). Which gave me the initial velocity of 15m/s. Lastly I did the d= 1/2 (velocity initial + velocity final)t equation which gave me a distance of 15 m. is this right?
 
  • #35
Finally after that when I replaced the distance i got into the work equation, I got 900J. Is this right
 
  • #36
You are impatient, aren't you? ;)

MsHomework said:
Ok well I decide to just do v=at. then I did velocity final= velocity initial + (acceleration)(distance). Which gave me the initial velocity of 15m/s. Lastly I did the d= 1/2 (velocity initial + velocity final)t equation which gave me a distance of 15 m. is this right?

What you write here is incorrect.
Surprisingly you do have the right answer for the distance.

The proper formula is:
distance = (velocity initial) (time) + (1/2) (acceleration) (time)^2.
MsHomework said:
Finally after that when I replaced the distance i got into the work equation, I got 900J. Is this right

Yes, this is right.

I was leading up to the trick that accelerating for 2 seconds from rest covers the same distance as decelerating in 2 seconds to rest.
I guess that won't be necessary any more.
 
  • #37
YUS! I UNDERSTAND NOW! Thank you for helping me! I will be posting another question soon, check it out!
 
  • #38
I hope you realize that velocity has to be in the direction the object is moving since it is the time rate of change of directed distance so no velocity cannot be in the direction opposite to its motion
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 56 ·
2
Replies
56
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
25
Views
3K