Solve Linear Momentum Problem: 6,000 N, 900 kg, 25 m, 20 m/s

In summary, you are trying to find the velocity of a car after it has started from rest and gained an initial speed of 20 m/s. If you use Newton's law, you will find that the car's final speed is 150,000 J.
  • #1
physicskillsme
24
0
I'm stuck with this.



The force from your car's motor is 6,000 N. The mass of the car is 900 kg. What will be you final speed after driving a distance of 25 m
a. if you start from rest
b. if you have an initial speed of 20 m/s.


I can get W=fd but I can't get how to find the speed (velocity)

W=fd

W=6000*25 = 150,000J

KE=1/2mv^2

I am at a loss where to pick up speed. I can get accleration by doing

A=F/m

A=6.67m/s/s

But accleration isn't the same as speed.

I know there must be a way of working back from the info I have but I really just can't see it.


I'd really appreciate some help.
 
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  • #2
Do you know the Kinematic equations of constant acceleration. Read FAQ. It is given there. For your information - Why do you think acceleration isn't constant here?
 
  • #3
vaishakh said:
Do you know the Kinematic equations of constant acceleration. Read FAQ. It is given there. For your information - Why do you think acceleration isn't constant here?


I'll take a llook at the FAQ then, but I'm still not sure what iit is I'm looking for.

Its not that I don't think accleration can't be constant but as far I know (am told by our teacher) that acceleration and speed are not the same thing. This is my first time in the Physics arena and I know this may all seem pretty simple but for me I can't make the obvious connections because I don't have prior knowledge.

Thanks for your reply. I'll see if I can get anything from the FAQ.
 
  • #4
You really do need kinematic equations to do this question.
 
  • #5
Ok I did a search for Kinematic equations of constant acceleration and nothing I can see that helps.

Clearly I need an equation - a pointer would be nice.

My situation is a good example of how students get turned of from the sciences, its not because they do not want to learn (I certainly do) but rather because when it comes to the simple stuff everyone expects you to say "oh yeah that's easy".

I think real world physics are important if not amazing but it seems to be a trial of errors and frustration when you start out because there seems to be an expectancy that we are born knowing the forumulas or something.

I don't mind doing the work but I need to know what the work is. Otherwise I could be wasting my time and setting myself up to fail.

So if anyoen could actually put my wheels on the right tracks and give me some links or info to follow I'd appreciate it. Telling me I should have an equation or understanding physics concepts is just resating the obvious and further killing any joy I might actually get from such an amazing subject.

My 2 cents
 
  • #6
Okay, firstly, you will need to use Newton's law:
[tex]F = ma[/tex]
to obtain the acceleration of the vehicle. Then you will need to use the kinematic equation:
[tex]v^2 = u^2 + 2as[/tex]
to obtain the final velocity. In the above equations [itex]F[/itex] is force, [itex]m[/itex] is mass, [itex]a[/itex] is acceleration, [itex]v[/itex] is final velocity, [itex]u[/itex] is initial velocity and [itex]s[/itex] is displacement.

Post your working if you want help or it checking :smile:

I would just like to say that it is better to learn new principles from a textbook, I would therefore recommend that you get yourself a good elementry physics textbook
 
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  • #7
Hootenanny said:
Okay, firstly, you will need to use Newton's law:
[tex]F = ma[/tex]
to obtain the acceleration of the vehicle. Then you will need to use the kinematic equation:
[tex]v^2 = u^2 + 2as[/tex]
to obtain the final velocity. In the above equations [itex]F[/itex] is force, [itex]m[/itex] is mass, [itex]a[/itex] is acceleration, [itex]v[/itex] is final velocity, [itex]u[/itex] is initial velocity and [itex]s[/itex] is displacement.

Post your working if you want help or it checking :smile:

I would just like to say that it is better to learn new principles from a textbook, I would therefore recommend that you get yourself a good elementry physics textbook

Ok thanks, I had the first part down but couldn't see the second formula and how to set it up (Which was really the problem) doing the work is easier. I have a class textbook but (I'll check) I don't recall seeing anything like the formula you have here. I was about to go off and try working with:

P=w/t =f*d/t then work P=f*d/t

Thus p=f*v and solve for v

where P=power w=work f=force d=displacement t=time. :confused:

Didnt get round to trying it out though.

Anyway what you have given here just from what little I know so I will try it out and see if the numbers make sense.


Oh and thanks for the help...sometimes having the answer helps understand the problem..it did here.
 
  • #8
Hootenanny said:
You really do need kinematic equations to do this question.

NO, you don't, just use conservation of energy and it will work out easily. Where does all that energy from work go? Into kinetic energy therefore just set work equal to kinetic energy.

In other words

W = Fs

W = (Delta)KE

So then

Fs = (Delta)KE
 
  • #9
d_leet said:
NO, you don't, just use conservation of energy and it will work out easily. Where does all that energy from work go? Into kinetic energy therefore just set work equal to kinetic energy.

In other words

W = Fs

W = (Delta)KE

So then

Fs = (Delta)KE

You are quite right, what I meant was it would be easier to answer both questions using kinematic equations.
 
  • #10
Ok for the record:

@Hootenay I used your formula and the question was considered answered correctly.

@D_Leet Your suggestion also worked and the instructor told me that this is what he was really looking for. The method you used would be true on all instances the first methos would only be tru if you were looking at a constant speed and were not concerned about any speed changes in between start and finish.

So I guess your both right and you can fight amongst yourselves over what the instructor has told me.

Whatever I would once again say thank you both for responding...I'll probably be back in the not too far distant future.
 
  • #11
physicskillsme said:
Ok for the record:
first methos would only be tru if you were looking at a constant speed and were not concerned about any speed changes in between start and finish.

This statement is incorrect, the kinematic equation is concerned with constant acceleration, which implies a change in velocity between the 'start' and 'finish'. Therefore, you can use my method for any instances where a constant acceleration is used.

You got the right answer, but make sure you know how :smile:
 
  • #12
Hootenanny said:
This statement is incorrect, the kinematic equation is concerned with constant acceleration, which implies a change in velocity between the 'start' and 'finish'. Therefore, you can use my method for any instances where a constant acceleration is used.

You got the right answer, but make sure you know how :smile:

Absolutely - so what do you think the instructor was getting at?
 
  • #13
physicskillsme said:
Absolutely - so what do you think the instructor was getting at?

I think you tutor meant/said Variable Acceleration, if this was the case the kinematic equation would be useless, you would have to use conservation of energy.
 
  • #14
OK well thanks for your input. I appreciate it.
 

What is linear momentum?

Linear momentum is a property of a moving object that describes its tendency to keep moving in a straight line at a constant speed. It is calculated by multiplying an object's mass by its velocity.

How do you solve a linear momentum problem?

To solve a linear momentum problem, you need to know the mass and velocity of an object, as well as the distance or time involved. Once you have this information, you can use the equation P = mv to calculate the linear momentum.

What are the units of linear momentum?

The units of linear momentum are kilogram meters per second (kg m/s). This is because momentum is calculated by multiplying mass (kg) by velocity (m/s).

What does the given information of "6,000 N, 900 kg, 25 m, 20 m/s" represent in a linear momentum problem?

The given information represents the force (6,000 N), mass (900 kg), distance (25 m), and velocity (20 m/s) involved in the problem. These values can be used to calculate the linear momentum of the object in question.

What are some real-world applications of solving linear momentum problems?

Linear momentum plays a role in many real-world applications, such as calculating the force needed for a car to stop in a certain distance, predicting the trajectory of a projectile, or understanding the impact of collisions in sports or engineering. It is also an important concept in understanding the motion of particles in physics and chemistry.

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