Calculating Force - Intuition vs. Solution

  • Thread starter Thread starter ohms law
  • Start date Start date
  • Tags Tags
    Force
Click For Summary
SUMMARY

The discussion focuses on calculating force in a physics problem involving an object with a mass of 68 kg propelled at a constant force for 3.0 seconds, resulting in a displacement of 2.25 m. The correct calculation yields a force of 34 N, derived from the acceleration of 0.50 m/s² using the formula F = ma. The participant expresses confusion regarding an alternative calculation that incorrectly suggests a force of 17 N, highlighting the importance of dimensional analysis and the role of dimensionless constants in physics equations.

PREREQUISITES
  • Understanding Newton's Second Law of Motion
  • Familiarity with basic kinematics
  • Knowledge of dimensional analysis
  • Concept of dimensionless constants in physics
NEXT STEPS
  • Study Newton's Second Law and its applications in various scenarios
  • Explore kinematic equations and their derivations
  • Learn about dimensional analysis and its significance in physics
  • Investigate the concept of dimensionless constants and their roles in physical equations
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of force calculations and dimensional analysis.

ohms law
Messages
70
Reaction score
0
I'm having a little trouble conceptualizing calculations of force. The problem statement was:
An object in space with a mass of 68 kg is propelled forward at a constant force (\vec{F}) for 3.0 seconds. After 3.0 s, the object has moved 2.25 m. find \vec{F}.

I can regurgitate the proper answer (34 N) by finding a_{x}=2 \Delta x/t^{2} = 0.50 m/s^{2} and \vec{F}=ma_{x} = (68kg)(0.50 m/s^{2}) = 34 N.

the problem is that my intuition tells me that it should be:
t = 3.0s
m = 68 kg
Δx = 2.25 m
So since N = kg \cdot m / s^{2}, N = 68 kg \cdot 2.25 m / 3.0 s^{2} = 17 N, which is obviously wrong. But, does that actually mean something else? Is that some sort of instantaneous value or something, or is it completely meaningless?
 
Physics news on Phys.org
What you have done is basically dimensional analysis. When you do this you need to keep in mind that the expression may contain a dimensionless constant, say 'k', which you need to include. Two quantities dimensionally equal aren't necessarily the same.

So, in your case it would be,

F \alpha [M][L][T]^{-2}

Therefore,

F = k * MLT^{-2}

Where the value of k is...? :wink:

...And, I feel good to be able to post again. :biggrin:
 

Similar threads

Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
19
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K