How Do You Calculate Work Done by a Non-Constant Force on a Variable Mass?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 5K views
madah12
Messages
326
Reaction score
1
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

Homework Statement


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= [tex]\int m*a dx[/tex] but in this case its m(t) and a(t) and x is also a function of t so how would the integral work?

Homework Equations


F=m*a
W=integral (ma)dx



The Attempt at a Solution

 
Physics news on Phys.org
Use the fact that:

[tex] a(t) = \frac{dv}{dt}[/tex]

to find the velocity - time equation:

[tex] v(t) = \int{a(t) \, dt} + C_{1}[/tex]

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

After you had found [itex]v(t)[/itex]. you simply use:

[tex] v(t) = \frac{dx}{dt} \Rightarrow dx = v(t) \, dt[/tex]

and substitute this into the integral for the total mechanical work. You integrate over t then.
 
Dickfore said:
Use the fact that:

[tex] a(t) = \frac{dv}{dt}[/tex]

to find the velocity - time equation:

[tex] v(t) = \int{a(t) \, dt} + C_{1}[/tex]

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

After you had found [itex]v(t)[/itex]. you simply use:

[tex] v(t) = \frac{dx}{dt} \Rightarrow dx = v(t) \, dt[/tex]

and substitute this into the integral for the total mechanical work. You integrate over t then.
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))
 
If you know [itex]a(t)[/itex] you need [itex]v(t_{0}) = v_{0}[/itex] to uniquely determine [itex]v(t)[/itex]. If you know [itex]x(t)[/itex], you can uniquely determine:

[tex] dx = \dot{x}(t) \, dt[/tex]

and you don't need additional information.
 
Dickfore said:
If you know [itex]a(t)[/itex] you need [itex]v(t_{0}) = v_{0}[/itex] to uniquely determine [itex]v(t)[/itex]. If you know [itex]x(t)[/itex], you can uniquely determine:

[tex] dx = \dot{x}(t) \, dt[/tex]

and you don't need additional information.

ok but like this I would need to integrate from t1 to t2 not from x1 to x2 right?
 
Yes! Your bounds must be with respect to whatever the [itex]d(\ldots)[/itex] quantity is in the integral.
 
ok so the question maybe seem strange but if I have a ball of ice which is being exposed to the sun light so the ice melts and it's moving in a straight line but is being pushed with a non constant force and I have x(t)(so also v(t) and a(t) , F(t) ,and I want to know the work done by the force from the time it is whole to where it melts what would be the general interface of the solution? should I get when m(t) is zero as dF(t)/da(t) =0?
(btw I am making the problem up)