Calculating Energy Dissipation for a Moving Object Using Integral Calculus

[Arm]

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

 

[TSny]

In the integral $\int v dx$, try switching to a time integration by letting $dx = \dfrac{dx}{dt}dt$.

 

[Arm]

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)

 

[erobz]

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.

 

[Arm]

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.

 

[erobz]

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.

$v = v(t) = \frac{d}{dt} \Big( x(t) \Big)$

 

[erobz]

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]

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.

 

[Arm]

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

 

[erobz]

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 $v(x) \neq v( x(t) )$?

 

[kuruman]

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?

 

[Arm]

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

 

[Arm]

You are welcome. Since you solved it, will you post the answer that you got?

$\frac{-ba^2T^5}{5}$

 

[SammyS]

$\dfrac{-ba^2T^5}{5}$

You might want to consider whether the result should be positive or negative.

 

[kuruman]

$\frac{-ba^2T^5}{5}$

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.