Work question without distance, work and velocity

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Homework Help Overview

The discussion revolves around a physics problem involving work done on a wagon due to an applied force opposing its motion. The scenario includes a wagon with a mass of 8.0 kg that is brought to rest by a force of 60 N over a duration of 2.0 seconds. Participants explore concepts related to work, force, momentum, and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between force, mass, and acceleration, with some suggesting the use of kinetic energy to determine work done. Questions arise regarding the interpretation of equations and the meaning of negative values for velocity. There is also exploration of different equations of motion and their applicability to the problem.

Discussion Status

The discussion is active, with participants offering various approaches to the problem. Some have provided guidance on using equations of motion, while others are clarifying concepts and definitions. There is a mix of interpretations and methods being explored, indicating a collaborative effort to understand the problem better.

Contextual Notes

Participants note the absence of distance in the original problem statement, which complicates the application of certain equations. The discussion also reflects on the implications of choosing positive and negative directions in the context of motion.

  • #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?
 
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  • #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
 

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