The work done by a variable force with respect to x distance is the integral of the function of that force with respect to x distance. But say we have a variable force as a function of time, can we say that the integral of the force with respect to time is equal to the work done by that variable force as a function of time?

 

Also, if I have the acceleration of a particle as a function of term and I have the mass of that particle, can I multiply the mass and acceleration to find a variable force as a function of time?

 

My intuition says yes to both question, but calculating the work using distance gives different results from calculating work using time.

Edited March 15, 201312 yr by Vay

The answer to the second question is yes, provided the mass is constant. But the answer to the first question is no. The integral of force with respect to time is impulse, not work. To find work, you must express force as a function of distance and integrate with respect to distance.

If you know F(t) and x(t) then you can find F(x), which is what you need to calculate work.

The integral of a force function of time w.r.t time, is the change in momentum of the object in which the force is applied, not the work done.
To do the integral, you'll need to find the relation between dx and dt then do substitution.

The work done by a variable force with respect to x distance is the integral of the function of that force with respect to x distance. But say we have a variable force as a function of time, can we say that the integral of the force with respect to time is equal to the work done by that variable force as a function of time?

 

Also, if I have the acceleration of a particle as a function of term and I have the mass of that particle, can I multiply the mass and acceleration to find a variable force as a function of time?

 

My intuition says yes to both question, but calculating the work using distance gives different results from calculating work using time.

 

The definition of mechanical work is

 

[math]W = \int \mathbf{F} d\mathbf{x}[/math]

 

which is valid for both variable and constant forces. Therein [math]\mathbf{F} = \mathbf{F}(\mathbf{x})[/math]. Using [math]\mathbf{x} = \mathbf{x}(t)[/math] and the definition of velocity [math]\mathbf{v} = d\mathbf{x}/dt[/math]

 

[math]W = \int \mathbf{F}(\mathbf{x}) d\mathbf{x} = \int \mathbf{F}(t) \mathbf{v} dt \neq \int \mathbf{F}(t) dt[/math]

 

Regarding the second question, the expression

 

[math]\mathbf{F} = m \mathbf{a}[/math]

 

is obtained from

 

[math]\mathbf{F} = \frac{d \mathbf{p}}{dt}[/math]

 

only when momentum is given by [math]\mathbf{p} = m \mathbf{v}[/math] and the mass is constant [math]dm/dt=0[/math]. Otherwise multiplying mass by the acceleration does not give the real force [math]\mathbf{F}[/math].

 

Therefore your intuitions are not correct.

Edited March 25, 201312 yr by juanrga