Energy dissipated by a force

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  • #1
Arm
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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
 

Answers and Replies

  • #2
TSny
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In the integral ##\int v dx##, try switching to a time integration by letting ##dx = \dfrac{dx}{dt}dt##.
 
  • #3
Arm
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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
erobz
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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.
 
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  • #5
Arm
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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
erobz
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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) ##
 
  • #7
erobz
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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.
 
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  • #8
kuruman
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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.
 
  • #9
Arm
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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.
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
erobz
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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) )##?
 
  • #11
kuruman
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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
Arm
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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
 
  • #13
Arm
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You are welcome. Since you solved it, will you post the answer that you got?
## \frac{-ba^2T^5}{5} ##
 
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  • #14
SammyS
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## \dfrac{-ba^2T^5}{5} ##
You might want to consider whether the result should be positive or negative.
 
  • #15
kuruman
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## \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.
 
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