Physics problem on derivatives and integrals

Question

So there are a few basic formulas I'd like to start with, $W=\int_0^bFdx$, $F=ma$, and $a=\frac{d^2}{dt^2}x$.

In words, Work $(W)$ is defined as the area under a Force versus Displacement $(F/x)$ graph, Force is defined mass times acceleration $(m\cdot a)$, and acceleration is defined as the second derivative of displacement with respect to time $(t)$.

While this may sound very simple, I am just confused on all of this.

Putting all of this together, I got

$$W=\int_0^bFdx=\int_0^bmadx$$

Assuming mass $(m)$ doesn't change,

$$W=m\int_0^badx$$

$$W=m\int_0^b\left(\frac{d^2}{dt^2}x\right)dx$$

At this point, I am slightly confused. How can I take the integral with respect to $x$ when $x$ is part of the definition of acceleration? An example situation and how to tackle it would be nice (assume acceleration is not constant).

I was wondering if it were possible to simplify the integral, given acceleration as a function of time.

Answer 1

The reason what you have there is tricky, is it usually results in a differential equation in $x$. so you can't in general take that integral. you might be able to find some identities, but that is about it.

Here is an example: If you want acceleration not constant, a classic example is a spring where you assume it follows Hook's law. So that's $F=-kx$ for some constant $k$. Just let k=1 for Simplicity. So you get $ma=-x$, so $\frac{d^2}{dt^2}x=-x/m$, this is a second order differential equation. IN this case it is pretty easy to find a solution, ( try looking it up, or finding one, if you don't already know), once you find it, it is not hard to take the integral. But the point here is, you can't do much with that integral you wrote in general, because acceleration could be almost any function of t.

Answer 2

When work is done by a variable force, one would calculate the area under the Force $(F(x))$ vs. Position $(x)$ graph.

So, the work integral would be

$$W=\int_{a}^b F(x) dx=\int_a^b m\cdot a(x) dx$$

Since $a(x)=\frac{dv(x)}{dt}=\frac{dv(x)}{dt}\frac{dx}{dt}=\frac{dv(x)}{dx}v(x)$, then $a(x)dx=v(x)dv(x)$

You can work with this integral, but you would still need to know the relevant information with respect to the position.

Answer 3

Even though you've accepted a previous answer, I wanted to expand a bit on vnd's comments to the question because they lead to a useful physical result.

Following his and Mackenzie's suggestions, let $v=\frac{da}{dt}$. The integral then becomes, per vnd's comments, $$W = m\int_{x_1}^{x_2}\left(\frac{d^2}{dt^2}x\right)\,dx = m\int_{v_1}^{v_2} v\,dv = \frac12mv_2^2-\frac12mv_1^2.$$ In other words, the net work is equal to the change in kinetic energy between the start and end points. You still need to find the velocity as a function of either time or position, but since you say that you have acceleration as a function of time, you can compute $$v(t) = v_0 + \int_0^t a(t)\,dt,$$ where $v_0$ is a known initial velocity.