Calculating Force with Mass and Velocity: A Quick Guide

  • Thread starter Thread starter Dejahboi
  • Start date Start date
  • Tags Tags
    Force Physics
Click For Summary
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.
Dejahboi
Messages
30
Reaction score
0

Homework Statement


M= 1320 kg in the (+)x-direction
V= 21.5 m/s
Find Force?

Homework Equations



F=MA

The Attempt at a Solution



(1320kg)(21.5 m/s) = 28380?
 
Physics news on Phys.org
1. Mass cannot have a direction. Did you mean the velocity?
2. You cannot solve the equation with the information you provided. Please use the homework help template provided to you. If this is from a textbook or homework assignement, copy the question exactly as it is given to you.
 
The question:

A 1320-kg car moves in the + x-direction with speed 21.5 m/s. Assuming constant braking and drag forces, find (a) the force and (b) the work needed to stop the car in a distance of 145 m.
 
Much better. Now we can solve it.
Before we start, though, you're supposed to take drag into consideration? Something seems wrong with that. Please write out the question exactly as it was given in your book or homework assignment.
 
That is exactly the question stated in the book :/. That's where I'm confused about too as well.
 
OK. We'll just go with the assumption that all the only forces acting on the care are gravity, the normal force of the road and a singular, constant force in the direction opposite the car's motion. You don't need to, but it would be good practice to draw a free body diagram of the car. You can check with the diagram that the only force we need to worry about is that constant force in the negative x direction.

You are given a constant force, and a constant mass, so there should be a constant acceleration. Try it from there.
 
I think the statement "Assuming constant braking and drag forces" means the rate of (-)acceleration and drag force are constant throughout the problem ie no impulse.
What are the initial and final velocities in the problem?
 
Impulse is the change of momentum. Since the car goes from 21.5 m/s to 0 m/s, there is a change in momentum, and therefore an impulse.

This is a badly worded problem, but I'm almost certain it's telling you that there is a single constant force, therefore constant acceleration. What are the equations of motion for constant acceleration?
 
I don't see a constant force being applied to the car? I drew the free body diagram and I did get the assumption there is no value in the x direction, hence that's what I'm looking for, but I honestly a bit stumped.
 
  • #10
What forces do you have acting on the car?
 
  • #11
-9.8N going up, 9.8 down, and 21.5 m/s in the + x-direction.
 
  • #12
What are the initial and final velocities? What is a change in velocity called?
 
  • #13
Initial would be 0, and final would be 21.5 m/s. Change of velocity is the acceleration
 
  • #14
Would the initial velocity be 0 or 21.5? Is there some equation that relates initial velocity, final velocity and acceleration?
 
  • #15
The 21.5m/s is a velocity, not a force. The velocity changes, so there has to be a force somewhere. The problem specifies a force F in the negative x direction. This gives you a constant a=F/m, also in the negative direction. What are the equations of motion with a constant acceleration?
 
  • #16
V=v0+Δat
 
  • #17
Is there an equation relating v, v0, a, distance?
 
  • #18
So I would use V2=V20+2a(X-X0)?
 
  • #19
There should be one that relates distance, initial velocity, final velocity, and acceleration, without time. If you don't know it, use x=x0+v0*t+1/2*a*t^2 and solve for t.
 
  • #20
You're missing a square on the left hand side.
 
  • #21
Yeah actually saw the typo after I posted the equation lol
 
  • #22
Ok, now what is v? v0? x? x0?
 
  • #23
Once you find the acceleration, what would you do?
 
  • #24
frogjg2003 said:
Ok, now what is v? v0? x? x0?

So V= 21.5m/s, V0= 0, X=145, and X0=0
jamesnb said:
Once you find the acceleration, what would you do?

I use F=ma?
 
  • #25
Yup, then how would you find work?
 
  • #26
You've got v and v0 mixed up. How fast is it going at the beginning of the problem? At the end of the problem?
 
  • #27
W=FΔx
 
  • #28
frogjg2003 said:
You've got v and v0 mixed up. How fast is it going at the beginning of the problem? At the end of the problem?

You're right, initial would be 21.5 m/s, and V final would be 0.
 
  • #29
Ok, now you've got everything you need to find the a. Multiply that by m, and you have F.
 
  • #30
You've got it.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
Replies
10
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
875
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
57
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K