Calculating Force with Mass and Velocity: A Quick Guide

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The discussion focuses on calculating the force required to stop a 1320 kg car moving at 21.5 m/s over a distance of 145 m. Participants clarify that mass does not have direction and emphasize the need to consider forces like braking and drag in the calculations. The correct approach involves using the equations of motion for constant acceleration to find acceleration and then applying F=ma to determine the force. There are discussions about the importance of keeping track of units and the definitions of impulse and acceleration. Ultimately, the conversation highlights the importance of careful problem interpretation and unit management in physics 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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