Particle Motion with Forces

In summary, the conversation is about solving a physics problem involving a conservative force acting on a 5.00 kg particle, with the equation F_x = (2x + 4) N. The task is to calculate the work done by the force (a), change in potential energy of the system (b), and kinetic energy of the particle at x=6.20m (c), given its speed of 3.00 m/s at x=1.40m. The conversation discusses integration, finding the area under the curve, and units of the calculated values. The expert suggests reviewing the definition of work and using a graph to understand the concept.
  • #1
brunie
62
0
Just need a bit of help with this question and to make sure I am solving it properly. Thanks
______________

A single conservative force acts on a 5.00 kg particle.
The equation F_x = (2x + 4) N describes this force, where x is in meters.
As the particle moves along the x-axis from x = 1.40 m to x = 6.20 m, calculate the following:

(a) work done by the force
(b) change in the potential energy of the system
(c) kinetic energy of the particle at x=6.20m
if its speed is is 3.00 m/s at x =1.40m

Ek = 0.5mv^2
Ep = mgh

***A)***

Ok I started by integrating the force as ∫ 2x +4
= x^2 + 4x
I then evaluated the difference in area between the two bounds, namely
((6.2)^2 + 4*6.2 - (1.4)^2 + 4*1.4) which gave 55.68
Since this was the force I multiplied by the change in distance 6.2 - 1.4 = 4.8
So 55.68 * 4.8 = 267.264

***B)***

I figured that since the particle is always on the x-axis, the potential energy difference should be zero.

***C)***

At 1.4m, speed is 3m/s.
Ek = 0.5(5)(3)^2
= 22.5

They want Ek at 6.2m.
And now I can't decide what to do. Because I think I need the speed at 6.2m in order to find the energy at 6.2m, but I don't kno why the Ek at 1.4m was given. Or I might need to use work and total energy. I am really not too sure.

Any help to verify this would be appreciated.
Thanks
 
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  • #2
For part A, the integral finds the area under the curve. What are the units of the area? (look at the units for the width of the interval times the units for the height or F_x value). Since the F_x value is force, and you're multiplying that by a distance, the units don't agree with what you think you found (a force.) You're definitely on the right track to take the integral, but I think you might see what I'm talking about if you look back at your work.

I agree with your part B.

Therefore, for part C, you started with a certain amount of kinetic energy when x was 1.40 meters. What happened to all the work that you did, since none of it was "changed" into potential energy (and the force was conservative)?
 
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  • #3
brunie said:
Just need a bit of help with this question and to make sure I am solving it properly. Thanks
______________

A single conservative force acts on a 5.00 kg particle.
The equation F_x = (2x + 4) N describes this force, where x is in meters.
As the particle moves along the x-axis from x = 1.40 m to x = 6.20 m, calculate the following:

(a) work done by the force
(b) change in the potential energy of the system
(c) kinetic energy of the particle at x=6.20m
if its speed is is 3.00 m/s at x =1.40m

Ek = 0.5mv^2
Ep = mgh

***A)***

Ok I started by integrating the force as ∫ 2x +4
= x^2 + 4x
I then evaluated the difference in area between the two bounds, namely
((6.2)^2 + 6.2 - (1.4)^2 + 1.4) which gave 41.28
Good :approve:
brunie said:
Since this was the force I multiplied by the change in distance 6.2 - 1.4 = 4.8
So 41.28 * 4.8 = 198.144
Not so good. What is the defintion of the work done by a force?
 
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  • #4
brunie said:
I then evaluated the difference in area between the two bounds, namely
((6.2)^2 + 6.2 - (1.4)^2 + 1.4) which gave 41.28

What happened to the 4? :confused:
 
  • #5
neutrino said:
What happened to the 4? :confused:
Good catch neutrino, I missed that ... :rolleyes:
 
  • #6
neutrino said:
What happened to the 4? :confused:

oops lol, let me fix that
 
  • #7
ok so for part A should I have multiplied again by the distance since it was integrated?
 
  • #8
Hootenany, why does the work have to go to a change in potential energy for part B? (Vs. a change in mechanical energy, of which there's another possibility in this problem)
 
  • #9
brunie said:
ok so for part A should I have multiplied again by the distance since it was integrated?
Lets just pause a moment and think. Write down for my the defintion of work done by a variable force F(x).
 
  • #10
brunie said:
ok so for part A should I have multiplied again by the distance since it was integrated?

1st, what do you multiply together to find work?
Then, draw a quick graph with Force on the vertical axis and displacement on the x-axis. Make a little rectangle somewhere on the graph. Now, if you were trying to find the area of that rectangle (using a formulas from geometry instead of an integral), what would you be multiplying together?
 
  • #11
drpizza said:
Hootenany, why does the work have to go to a change in potential energy for part B? (Vs. a change in mechanical energy, of which there's another possibility in this problem)
Oops edit, my bad, misread the question, for some reason I thought we were in the vertical plane :redface:
 
  • #12
Hootenanny said:
Lets just pause a moment and think. Write down for my the defintion of work done by a variable force F(x).

Isnt work just the force by the distance. And since the integrated area gave the total force in Nm shouldn't it just be that?
 
  • #13
brunie said:
Isnt work just the force by the distance. And since the integrated area gave the total force in Nm shouldn't it just be divided by the distance in metres twice for the work in N/m?
No, not quite. The defintion of work done (W) by a variable force (F) is gived by;

[tex]W = \int^{x_1}_{x_0}F(x)dx[/tex]
 
  • #14
Hootenanny said:
No, not quite. The defintion of work done (W) by a variable force (F) is gived by;

[tex]W = \int^{x_1}_{x_0}F(x)dx[/tex]

By integrating that wouldn't it just be an extra x term, meaning that its divided by the distance to get rid of it?
 
  • #15
brunie said:
By integrating that wouldn't it just be an extra x term, meaning that its divided by the distance?
:bugeye: I don't quite follow your logic there. Let's take your function above;

[tex]W = \int^{x_1}_{x_0} (2x+4)dx = \left[ x^2 + 4x \right]^{x_1}_{x_0}[/tex]

[tex]W = \left(x_1^2+4x_1\right)-\left(x_0^2+4x_0\right)[/tex]
 
  • #16
Hootenanny said:
:bugeye: I don't quite follow your logic there. Let's take your function above;

[tex]W = \int^{x_1}_{x_0} (2x+4)dx = \left[ x^2 + 4x \right]^{x_1}_{x_0}[/tex]

[tex]W = \left(x_1^2+4x_1\right)-\left(x_0^2+4x_0\right)[/tex]

ok, i evaluated that already but I am just confused about what to do with that result
 
  • #17
brunie said:
ok, i evaluated that already but I am just confused about what to do with that result
If you evaluated the integral correctly, it will have returned the value of the work done by the force as the particle moved between x0 and x1
 
  • #18
Hootenanny said:
If you evaluated the integral correctly, it will have returned the value of the work done by the force as the particle moved between x0 and x1

so by integrating the equation and evaluating w.r.t the endpoints i am finding the work done by the force?
 
  • #19
brunie said:
Isnt work just the force by the distance. And since the integrated area gave the *answer* in Nm shouldn't it just be that?
Fixed.
Yep. Newton-meters are not a unit of force.

Maybe it'll help to go back to something simpler.

For a similar problem, let's do this:
On the vertical axis, put velocity. On the horizontal axis, let's put time. Now, here's an equation. v_t= 5.
Find the displacement from t=1 to t=7.
Note: if you graph this, it's just a constant function (a horizontal line).

If we integrate, we'd have [tex]displacement = \int^{7}_{1}5dt[/tex]
This integral is 5t evaluated from 1 to 7, or 5*7 - 5*1 = 30 (units??)
Now instead, notice that it's just a rectangle. You can find the area in a rectangle by a simple length * width. The length is 6 seconds (from 1 second to 7 seconds) and the width is 5 m/s. Thus, you're multiplying meters/second by seconds, and the units cancel to give you an answer in meters.

The area under a velocity vs time curve is the displacement.
Likewise, the area under a Force vs. distance curve IS the work.
Thus, as Hootenanny said, Work = the integral...
(the integral is just a way to find the area under a curve.)
 
  • #20
brunie said:
so by integrating the equation and evaluating w.r.t the endpoints i am finding the work done by the force?

Yes
That *is* the work. There's no need to multiply by anything.

(Don't you wish we would have just told you that you went too far on your first step the first time you did it? :tongue2: ) Sometimes it's tough to help someone understand the process and what they're actually doing vs. just rote memorization of steps that you might end up confusing later on.
 
  • #21
great, thanks for clarifying that
i was beggining to get very confused

so we kno the work is the integrated equation evaluated and that the change in potential is zero (since it is on the x-axis),

but for C)
since we kno the work is 55.68
and that Ek at 1.4m is 22.5
for Ek at 6.2 is it simply 55.68 - 22.5 = 33.18
?
 
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  • #22
brunie said:
great, thanks for clarifying that
i was beggining to get very confused

so we kno the force is the integrated equation evaluated and that the change in potential is zero (since it is on the x-axis),

but for C)
since we kno the total force is 55.68
and that Ek at 1.4m is 22.5
for Ek at 6.2 is it simply 55.68 - 22.5 = 33.18
?

You did it again! No, the force isn't the integrated equation evaluated. The *WORK* is the integrated equation evaulated.
 
  • #23
drpizza said:
You did it again! No, the force isn't the integrated equation evaluated. The *WORK* is the integrated equation evaulated.

o you ya sry i actually meant to put work, ill fix it
 
  • #24
You're on the right track. Now, you did work on the car (or whatever it was), by exerting a force in the positive direction. The car was originally moving in the positive direction. Would you want to add the energy (from the work) to the kinetic energy? Or do you want to subtract?
 
  • #25
drpizza said:
You're on the right track. Now, you did work on the car (or whatever it was), by exerting a force in the positive direction. The car was originally moving in the positive direction. Would you want to add the energy (from the work) to the kinetic energy? Or do you want to subtract?

im not too sure because wouldn't the initial kinetic be a part of the total work while the total kinetic would equal the work

so W = Ek1 + Ek2
and then Ek2 would be W - Ek1

??
 
  • #26
The work done on the system happened when it already had a kinetic energy. Let's look at it this way. If you had a cart that started at rest (the velocity at x = 1.4 was 0 meters/second), and you still had the same force function, you would have done 55.68 joules of work on that car. Thus, the car would now have 55.68 joules of kinetic energy (if it started from rest.)

But, the car didn't start from rest. It already had some kinetic energy. After doing the same amount of work, (and pushing it in the same direction that it's already going), would you expect it to have more kinetic energy than the car that was initially at rest, or less kinetic energy than the car that was initially at rest?
 
  • #27
drpizza said:
The work done on the system happened when it already had a kinetic energy. Let's look at it this way. If you had a cart that started at rest (the velocity at x = 1.4 was 0 meters/second), and you still had the same force function, you would have done 55.68 joules of work on that car. Thus, the car would now have 55.68 joules of kinetic energy (if it started from rest.)

But, the car didn't start from rest. It already had some kinetic energy. After doing the same amount of work, (and pushing it in the same direction that it's already going), would you expect it to have more kinetic energy than the car that was initially at rest, or less kinetic energy than the car that was initially at rest?

ok that analogy helps the situation make sense
so its 22.5 + 55.68 = 78.18
thanks a lot for ur help, it was really appreciated
 
  • #28
You're welcome. Good luck!
 

1. What is particle motion with forces?

Particle motion with forces is the study of how forces affect the movement of particles. It involves understanding the relationship between forces acting on a particle and the resulting motion of that particle.

2. What are the different types of forces that can affect particle motion?

There are several types of forces that can affect particle motion, including gravitational forces, electromagnetic forces, and contact forces such as friction and tension.

3. How do you calculate the net force on a particle?

The net force on a particle is calculated by adding all the forces acting on the particle together. This can be done by using vector addition, where the magnitude and direction of each force are taken into account.

4. What is Newton's Second Law of Motion?

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be written as the equation F=ma, where F is the net force, m is the mass, and a is the acceleration.

5. How does the direction of a force affect particle motion?

The direction of a force is an important factor in determining the resulting motion of a particle. For example, if a force is applied in the same direction as the particle's motion, it will increase the particle's speed. However, if the force is applied in the opposite direction, it will slow down the particle's motion.

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