Calculating Force with Mass and Velocity: A Quick Guide

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Homework Help Overview

The discussion revolves around calculating the force acting on a car with a given mass and velocity, specifically in the context of braking forces and work done to stop the car over a specified distance. The subject area includes dynamics and kinematics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between mass, velocity, and force, questioning the initial setup of the problem and the assumptions regarding forces acting on the car. There are discussions about the implications of constant acceleration and the definitions of impulse and momentum.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to approach the problem, including the importance of free body diagrams and the equations of motion. There is an acknowledgment of confusion regarding the problem's wording and the need for clarity on initial and final velocities.

Contextual Notes

Participants note the importance of considering all forces acting on the car, including drag and braking forces, and the need to clarify the problem's assumptions. There are also reminders about the significance of units in calculations.

  • #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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