# Work by a non constant force

1. Jul 25, 2010

I don't know that if this is the right place to post this but it's safer than to post it in the general physics and get a warning
1. The problem statement, all variables and given/known data
let say we have an object moving with a non constant acceleration and has a non constant mass where we have a(t) as any function like t^3 or (7t^2-3)/2t and m(t) also is a function of time and of course position changes with respect to time I know the formula for work is W= $$\int m*a dx$$ but in this case its m(t) and a(t) and x is also a function of t so how would the integral work?

2. Relevant equations
F=m*a
W=integral (ma)dx

3. The attempt at a solution

2. Jul 25, 2010

### Dickfore

Use the fact that:

$$a(t) = \frac{dv}{dt}$$

to find the velocity - time equation:

$$v(t) = \int{a(t) \, dt} + C_{1}$$

where $C_{1}$ is an arbitrary integrating constant which can be determined only if you know the velocity at a particular instant in time $v(t_{0}) = v_{0}$. Judging by what you had given in the statement of the problem, I do not see such kind of information.

After you had found $v(t)$. you simply use:

$$v(t) = \frac{dx}{dt} \Rightarrow dx = v(t) \, dt$$

and substitute this into the integral for the total mechanical work. You integrate over t then.

3. Jul 25, 2010

so in this case to calculate the work it would be necessary to know the velocity?
( I am assuming the information given is only a(t) and m(t) or F(t) and x(t))

4. Jul 25, 2010

### Dickfore

If you know $a(t)$ you need $v(t_{0}) = v_{0}$ to uniquely determine $v(t)$. If you know $x(t)$, you can uniquely determine:

$$dx = \dot{x}(t) \, dt$$

and you don't need additional information.

5. Jul 25, 2010

ok but like this I would need to integrate from t1 to t2 not from x1 to x2 right?

6. Jul 25, 2010

### Dickfore

Yes! Your bounds must be with respect to whatever the $d(\ldots)$ quantity is in the integral.

7. Jul 25, 2010