Calculating Force with Mass and Velocity: A Quick Guide

[Dejahboi]

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?

 

[frogjg2003]

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.

 

[Dejahboi]

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.

 

[frogjg2003]

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.

 

[Dejahboi]

That is exactly the question stated in the book :/. That's where I'm confused about too as well.

 

[frogjg2003]

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.

 

[jamesnb]

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?

 

[frogjg2003]

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?

 

[Dejahboi]

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.

 

[frogjg2003]

What forces do you have acting on the car?

 

[Dejahboi]

-9.8N going up, 9.8 down, and 21.5 m/s in the + x-direction.

 

[jamesnb]

What are the initial and final velocities? What is a change in velocity called?

 

[Dejahboi]

Initial would be 0, and final would be 21.5 m/s. Change of velocity is the acceleration

 

[jamesnb]

Would the initial velocity be 0 or 21.5? Is there some equation that relates initial velocity, final velocity and acceleration?

 

[frogjg2003]

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?

 

[Dejahboi]

V=v₀+Δat

 

[jamesnb]

Is there an equation relating v, v0, a, distance?

 

[Dejahboi]

So I would use V²=V²₀+2a(X-X₀)?

 

[frogjg2003]

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.

 

[frogjg2003]

You're missing a square on the left hand side.

 

[Dejahboi]

Yeah actually saw the typo after I posted the equation lol

 

[frogjg2003]

Ok, now what is v? v0? x? x0?

 

[jamesnb]

Once you find the acceleration, what would you do?

 

[Dejahboi]

Ok, now what is v? v0? x? x0?

So V= 21.5m/s, V₀= 0, X=145, and X₀=0

Once you find the acceleration, what would you do?

I use F=ma?

 

[jamesnb]

Yup, then how would you find work?

 

[frogjg2003]

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?

 

[Dejahboi]

W=FΔx

 

[Dejahboi]

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.

 

[frogjg2003]

Ok, now you've got everything you need to find the a. Multiply that by m, and you have F.

 

[jamesnb]

You've got it.

 

[frogjg2003]

Make sure you're watching the signs. v<v0, and x>x0, so a should be negative. Therefore so should the force and work.

 

[Dejahboi]

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

 

[frogjg2003]

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?

 

[Dejahboi]

N/kg = J :) My mistake, therefore the kg cancel :)

 

[jamesnb]

Don't forget they asked for force to stop the car and work done by the brakes.

 

[Dejahboi]

:(, oh I forgot about part B, alright back to it again

 

[Dejahboi]

Okay so

(-2104 J) (145m) = -305080 J?

 

[jamesnb]

frogjg, going back to impulse and acceleration, isn't impulse the first derivative of acceleration? Meaning if acceleration is constant, there is no impulse?

 

[jamesnb]

Dejahboi, that's what I got.

 

[Dejahboi]

Thanks!

 

[frogjg2003]

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.

 

[jamesnb]

Except mind your units. The force is in Newtons.

 

[frogjg2003]

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.

 

[Dejahboi]

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!

 

[frogjg2003]

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.