Force on 500g Particle: Velocity at t = 2s

  • Thread starter Thread starter bestchemist
  • Start date Start date
  • Tags Tags
    Force Particle
Click For Summary

Homework Help Overview

The problem involves a 500g particle subjected to a time-varying force, described by the equation Fx = (4−(t/s)²) N, and seeks to determine the particle's velocity at t = 2 s. The initial velocity at t = -2 s is given as vx = -4.0 m/s. Participants are exploring the implications of using different equations of motion under non-constant force conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the applicability of equations of motion when the force is not constant, questioning the use of specific formulas. Some express uncertainty about the calculations leading to the proposed velocity of 4 m/s, while others suggest considering the average force or the relationship between velocity and force as functions of time.

Discussion Status

The discussion is ongoing, with participants raising questions about the correct approach to take given the non-constant nature of the force. Some guidance has been offered regarding the need to calculate the average force, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants are grappling with the implications of using equations that assume constant force, and there is a noted confusion regarding the units and validity of the formulas applied. The original poster's calculations have been questioned, highlighting the need for clarity in the approach to the problem.

bestchemist
Messages
37
Reaction score
0
A 500g particle has velocity vx = -4.0m/s at t = - 2 s. Force Fx = (4−(t/s)2) N is exerted on the particle between t = - 2 s and t = 2 s. This force increases from 0 N at t = - 2 s to 4 N at t = 0 s and then back to 0 N at t = 2 s. What is the particle’s velocity at t = 2 s?
I'm not sure if the answer is 4 m/s. can someone check if I do it right?

V2 = V02 + at
v= v0+ (F/m)t
v = -4m/s + (0/0.5)*2
v= 4m/s
 
Physics news on Phys.org
v= v0+ (F/m)t requires a constant force, or you have to calculate the average force. You cannot use the force at some arbitrary point in time here.

The first equation looks odd.
How did you get the last line from the line before?
 
It doesn't have constant force so which formula do I use?

v = 4 m/s?
I got that from plug in v02
v0 = -4 m/s
so I thought it would be 4 m/s since I have to take square root to get v
Is it wrong? lol
 
bestchemist said:
It doesn't have constant force so which formula do I use?
How are the velocity as function of time and the force related?

Alternatively, calculate the average force.

I got that from plug in v02
v0 = -4 m/s
The first formula was wrong anyway. You can see it if you check the units.

so I thought it would be 4 m/s since I have to take square root to get v
Is it wrong? lol
v2=16m2/s2 has two solutions: v=4m/s and v=-4m/s. But the formula you used was not right anyway.
 
I figure it out! Thank you :)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
2K