Calculating Force with Mass and Velocity: A Quick Guide

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SUMMARY

The discussion focuses on calculating the force required to stop a 1320 kg car moving at 21.5 m/s in the positive x-direction over a distance of 145 m, using the equation F=MA. Participants clarify that the problem requires considering constant braking and drag forces, leading to a constant acceleration. The final calculations yield a force of -2104 N and work done of -305080 J, emphasizing the importance of unit consistency in physics calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F=MA)
  • Knowledge of kinematic equations for constant acceleration
  • Familiarity with the concepts of force, work, and energy
  • Ability to perform unit conversions and dimensional analysis
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  • Study kinematic equations for constant acceleration in-depth
  • Learn about the relationship between force, mass, and acceleration
  • Explore the concept of impulse and its relation to momentum
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Students in physics courses, educators teaching mechanics, and anyone interested in understanding the principles of force and motion in real-world applications.

  • #31
Make sure you're watching the signs. v<v0, and x>x0, so a should be negative. Therefore so should the force and work.
 
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  • #32
OMG, thank you!

So,

0=21.52+2a(145-0)

Set it equal to a,

a= -462.25/290

a= -1.59

F=ma

F= (1320kg)(-1.59N/kg)

=-2104 J
 
  • #33
The numbers are right, but the units are wrong. What is the unit of force (name and base units)?
What are the base units for J?
 
  • #34
N/kg = J :) My mistake, therefore the kg cancel :)
 
  • #35
Don't forget they asked for force to stop the car and work done by the brakes.
 
  • #36
:(, oh I forgot about part B, alright back to it again
 
  • #37
Okay so

(-2104 J) (145m) = -305080 J?
 
  • #38
frogjg, going back to impulse and acceleration, isn't impulse the first derivative of acceleration? Meaning if acceleration is constant, there is no impulse?
 
  • #39
Dejahboi, that's what I got.
 
  • #40
Thanks!
 
  • #41
That's not right either.

velocity is distance per time: m/s
acceleration is velocity per time: m/s/s=m/s^2
force is mass times acceleration: N=kg*m/s^2
work is force times distance: J=N*m=(kg*m/s^2)*m=kg*m^2/s^2

Keeping track of base units can sometimes be a pain, but you have to work through it.
 
  • #42
Except mind your units. The force is in Newtons.
 
  • #43
jamesnb said:
frogjg, going back to impulse and acceleration, isn't impulse the first derivative of acceleration? Meaning if acceleration is constant, there is no impulse?

I was taught that impulse was the integral of force, or the change in momentum. Jerk is the derivative of acceleration. If you want to continue this discussion, you can pm me. I don't want to clog up this thread with a tangent that will only serve to distract/confuse Dejahboi.
 
  • #44
frogjg2003 said:
That's not right either.

velocity is distance per time: m/s
acceleration is velocity per time: m/s/s=m/s^2
force is mass times acceleration: N=kg*m/s^2
work is force times distance: J=N*m=(kg*m/s^2)*m=kg*m^2/s^2

Keeping track of base units can sometimes be a pain, but you have to work through it.

Thanks! I'll make note in my notebook with this info. This will help me with my next problem :). THanks again!
 
  • #45
The internet is your friend. Wikipedia has pages for pretty much every unit imaginable. Google calculator can handle units. There is a site: wolframalpha.com that is basically a limited version of Mathematic, an analytic mathematics program. This is the age of the internet and you should take full advantage. Joining PF was just the beginning.
 

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