Calculating Energy Dissipation for a Moving Object Using Integral Calculus

In summary, the conversation discusses the process of finding an integral involving the function v(x) and the force F, which is a function of velocity. The integral is solved using the substitution method by changing the variable of integration from x to t, resulting in two possible solutions: ##\int F dx = -b \int v(t) \cdot v(t) dt## or ##\int F dx = -b \int v(x) dx##. The latter solution is preferred as it can be easily solved by hand.
  • #1
Arm
16
5
Homework Statement
Question
An object moves in one dimension according to the function ##x(t) = \frac{1}{3}at^3 ##, where a is a positive constant with units of ##\frac{m}{s^3}.## During this motion, a force F =−bv is exerted on the object, where v is the object’s velocity and b is a constant with units of ##\frac{kg}{s}. ## Which of the following expressions will yield the amount of energy dissipated by this force during the time interval from t=0 and t=T ?
Relevant Equations
##ΔE = \int_{x_1}^{x_2} F(x) dx ##
##F = -b*v##
##x(t) = \frac{1}{3}at^3 ##
##v(t) = at^2##
Since ##F = -b * v## I said ## F(x) = -b*v(x) ##
Also x(0) = 0 and x(T) = ## \frac{1}{3}aT^3 ##

I made the integral ## ΔE = -b* \int\limits_0^\frac{aT^3}{3} v(x) dx ##

I wanted to replace v(x) with v(t) somehow
I first tried ##v(x) = ax^2## but then realized that was wrong
Then I tried using ## x = v*t ## but couldn't do anything with it
 
Physics news on Phys.org
  • #2
In the integral ##\int v dx##, try switching to a time integration by letting ##dx = \dfrac{dx}{dt}dt##.
 
  • Like
Likes topsquark
  • #3
TSny said:
In the integral ##\int v dx##, try switching to a time integration by letting ##dx = \dfrac{dx}{dt}dt##.
That would make it ##\int v(x)*v(t) dt##, I don't know what to do from there. (I've only done a handful of integrals before)
 
  • #4
Arm said:
That would make it ##\int v(x)*v(t) dt##, I don't know what to do from there. (I've only done a handful of integrals before)
You don't have ##v(x)## in the integrand. You have ##v(t)## by differentiating ##x(t)## w.r.t time ##t##. The Force ##F## is a function of velocity, which is a function of time. That means ##F = f(t)##. While it does vary with position, it does so by varying the function for ##v## in time.
 
Last edited:
  • Like
Likes topsquark
  • #5
erobz said:
You don't have ##v(x)##. You have ##v(t)## by differentiating ##x(t)## w.r.t time ##t##.
So is ##v(x) = \dfrac{d(x(t))}{dt}##? I'm confused right now about how to represent v(x) as a derivative.
 
  • #6
Arm said:
So is ##v(x) = \dfrac{d(x(t))}{dt}##? I'm confused right now about how to represent v(x) as a derivative.
Where are you coming up with ##v(x)##? You don't need it. You have changed the variable of integration from ##x## to ##t##, with the substitution @TSny showed.

##v = v(t) = \frac{d}{dt} \Big( x(t) \Big) ##
 
  • Like
Likes topsquark
  • #7
If you want to do it the long way solve for ##t(x)##. Then substitute that into ##v(t)##. $$v(x) = v( t(x) )$$ Then, the integral is:

$$ \int F dx = -b \int v( t(x) ) dx = - b \int_{x(0)}^{x(T)} v(x) dx $$

But I think you will find it preferable to transform the integral to the time domain.

$$ \int F dx = \int F(t) dx = -b \int v(t) dx = -b \int v(t) \cdot v(t) dt = -b \int_{0}^{T} {v(t)}^2 dt $$

You should do both so you can play around with these concepts.
 
Last edited:
  • Like
Likes topsquark
  • #8
To @Arm:
Look,
If ##x = \frac{1}{3}at^3##, it follows that ##v=at^2.##
Following @TSny's suggestion,
##Fdx=F\dfrac{dx}{dt}dt=F(at^2)dt = (-bat^2)(at^2)dt##
Integrate.
 
  • Like
Likes topsquark
  • #9
erobz said:
If you want to do it the long way solve for ##t(x)##. Then substitute that into ##v(t)##. $$v(x) = v( t(x) )$$ Then, the integral is:

$$ \int F dx = -b \int v( t(x) ) dx = - b \int_{x(0)}^{x(T)} v(x) dx $$

But I think you will find it preferable to transform the integral to the time domain.

$$ \int F dx = \int F(t) dx = -b \int v(t) dx = -b \int v(t) \cdot v(t) dt = -b \int_{0}^{T} {v(t)}^2 dt $$

You should do both so you can play around with these concepts.
kuruman said:
To @Arm:
Look,
If ##x = \frac{1}{3}at^3##, it follows that ##v=at^2.##
Following @TSny's suggestion,
##Fdx=F\dfrac{dx}{dt}dt=F(at^2)dt = (-bat^2)(at^2)dt##
Integrate.
My calculus is very bad which is why I don't follow the steps where you change domains, but I ended up solving it by using ##v(x(t))##

Thank you
 
  • #10
Arm said:
My calculus is very bad which is why I don't follow the steps where you change domains, but I ended up solving it by using ##v(x(t))##

Thank you
I hope you mean you used ##v(x) = v( t(x) )##, because ##v(x) \neq v( x(t) )##?
 
  • Like
Likes topsquark
  • #11
Arm said:
My calculus is very bad which is why I don't follow the steps where you change domains, but I ended up solving it by using ##v(x(t))##

Thank you
You are welcome. Since you solved it, will you post the answer that you got?
 
  • #12
erobz said:
I hope you mean you used ##v(x) = v( t(x) )##, because ##v(x) \neq v( x(t) )##?
Yeah that's what I meant, mistyped the reply
 
  • Like
Likes erobz
  • #13
kuruman said:
You are welcome. Since you solved it, will you post the answer that you got?
## \frac{-ba^2T^5}{5} ##
 
  • Like
Likes SammyS and erobz
  • #14
Arm said:
## \dfrac{-ba^2T^5}{5} ##
You might want to consider whether the result should be positive or negative.
 
  • #15
Arm said:
## \frac{-ba^2T^5}{5} ##
Thank you for posting your solution. Note that if you followed the method in post #8, you could have done it in your head, $$\int_0^T(-bat^2)(at^2)dt=-\frac{1}{5}ba^2T^5.$$It's something worth considering for future reference next time you encounter a problem like this.
 
Last edited:
  • Like
Likes Arm

Related to Calculating Energy Dissipation for a Moving Object Using Integral Calculus

What is the definition of energy dissipated by a force?

Energy dissipated by a force is the amount of energy that is transformed into a different form or lost as heat due to the application of a force on an object.

How is energy dissipated by a force calculated?

The energy dissipated by a force can be calculated by multiplying the magnitude of the force by the distance over which the force is applied. This is known as the work-energy theorem.

What factors affect the amount of energy dissipated by a force?

The amount of energy dissipated by a force is affected by the magnitude of the force, the distance over which the force is applied, and the properties of the object on which the force is applied (such as its mass and composition).

What happens to the energy dissipated by a force?

The energy dissipated by a force is typically converted into different forms of energy, such as thermal energy or sound energy. It can also be stored as potential energy, such as in a compressed spring or stretched rubber band.

Can the energy dissipated by a force be recovered?

In most cases, the energy dissipated by a force cannot be fully recovered. Some of the energy is lost as heat, which is difficult to convert back into usable energy. However, there are some methods, such as regenerative braking, that can partially recover the energy dissipated by a force.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
135
  • Introductory Physics Homework Help
Replies
6
Views
445
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
714
  • Introductory Physics Homework Help
Replies
3
Views
926
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
725
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
267
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top