Deriving Work-Kinetic Energy Theorem

Question

I am currently reading Physics for Scientists and Engineers (Ninth Edition) by Serway and Jewett and in Chapter 7.5, a derivation of the work-kinetic energy theorem was shown.

To give context, consider a system consisting of an object of mass $m$ moving through a displacement directed to the right due to a net force $\sum F$ , also directed to the right.

The derivation process was then, $$\begin{aligned} W_{\text{ext}} &= \int_{x_i}^{x_f} \sum F \ dx \\ &= \int_{x_i}^{x_f} ma \ dx \\ &= \int_{x_i}^{x_f} m \frac{dv}{dt} \ dx \\ &= \int_{x_i}^{x_f} m \frac{dv}{dx} \frac{dx}{dt} \ dx \\ &= \int_{v_i}^{v_f} mv \ dv \end{aligned}$$

Unfortunately, I'm having trouble understanding how the author arrived at the final step from the previous one.

I have tried equating $v = dx/dt$ and using substitution rule with the same variable $v = x$ and $dv = dx$ to arrive at

$$W_{\text{ext}}= \int_{v_i}^{v_f} mv \frac{dv}{dv} \ dv$$

and then cancelling the $dv$'s as if they were a fraction to get the same result as the book but I think I'm terribly mistaken.

Can someone perhaps shed insight on how to proceed with this problem?

Answer 1

A better way if you're not comfortable with the method shown:

$$\frac{dx}{dt} = v$$

$$dx = v dt$$

$$W = \int_{a}^{b} F dx$$ $$\int_{t_{0}}^{t} [m \frac{dv}{dt}] [v dt]$$

Rewriting:

$$m \int_{t_{0}}^{t} [v \frac{dv}{dt}] dt$$

Instead of "cancelling" dt, simply use the inverse chain rule to integrate this expression with respect to time directly, (raise by the power, divide by that power, divide by the derivative of the inside function!)

or in reverse : $\frac{d}{dt} \frac{1}{2}v^2 = v \frac{dv}{dt}$ $$W = m [\frac{1}{2} v(t)^2 - \frac{1}{2} v(t_{0})^2]$$

Answer 2

$$\begin{aligned} W_{\text{ext}} &= \int_{x_i}^{x_f} \sum F \ dx \\ &= \int_{x_i}^{x_f} ma \ dx \\ &= \int_{x_i}^{x_f} m \frac{dv}{dt} \ dx \end{aligned}$$ From this point, we note that $x = x(t)$, so: $$\int_{x_i}^{x_f} m \frac{dv}{dt} \ dx = \int_{t_i}^{t_f} m \frac{dv}{dt} \ \frac{dx}{dt} \ dt = \int_{t_i}^{t_f} m \frac{dv}{dt} \ v \ dt$$

Here we can use integration by parts, where both functions are $v$: $$\int_{t_i}^{t_f} m v \frac{dv}{dt} \ dt = m(vv)_{t_i}^{t_f} - \int_{t_i}^{t_f} m v\frac{dv}{dt} \ dt$$ Because the left and right integral are the same: $$\int_{t_i}^{t_f} m v \frac{dv}{dt} \ dt = \Delta \left(\frac{1}{2}m v^2 \right)$$

Answer 3

From the second step, $$ma.dx = m.\frac{dv}{dt}.dx = m.dv.\frac{dx}{dt}$$ And with $\frac{dx}{dt} = v$, the integrand becomes $mv.dv$, as written in the last step.

Your substitution $v = x$ does not make sense as $v$ is defined to be $\frac{dx}{dt} ≠ x$, which is why it does not work.

Hope this helps.

Answer 4

You have overcomplicated your derivation by trying to derive within the integral. Start with the differential forms first then integrate at the end

$$W = Fdx$$ $$W = m \frac{dv}{dt} dx$$ $$W = m \frac {dx}{dt} dv$$ $$W = mvdv$$

Answer 5

We start with the definition of work: $$W = \int_{x_{i}}^{x_{f}} F\cdot dx$$ From Newton's second law: $$W = \int_{x_{i}}^{x_{f}} m\frac{dv}{dt}\cdot dx$$ $$W = \int_{t_{i}}^{t_{f}} m\frac{dv}{dt}\cdot vdt$$ $$W = \underbrace{\int_{t_{i}}^{t_{f}}\left( m\frac{dv}{dt}\cdot v\right)dt}_{I}$$

We use integration by parts: $$\int f'g=fg-\int fg'$$

$$W = m v^2\biggr|_{t_{i}}^{t_{f}} - \underbrace{\int_{t_{i}}^{t_{f}}\left( mv \cdot\frac{dv}{dt}\right)dt}_{I}$$

From this we get that: $$W=\frac{1}{2}m\left[v\left(t_{f}\right)^2-v\left(t_{i}\right)^2\right]$$

Answer 6

\begin{equation} a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt} \end{equation} via chain rule, therefore \begin{equation} a=\frac{dv}{dx}v=v\frac{dv}{dx} \ \xrightarrow{} \ a\,dx= v\, dv \end{equation} Now you should be able to understand.