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

AI Thread Summary
To calculate work done by a non-constant force on a variable mass, the integral for work must account for time-dependent mass and acceleration. The formula W = ∫ m(t) * a(t) dx requires knowing the velocity function v(t), which can be derived from acceleration a(t) by integrating it over time. The relationship dx = v(t) dt allows for substituting into the work integral, necessitating integration with respect to time rather than position. If additional information about initial conditions, such as velocity or position, is provided, it can uniquely determine the velocity function. Ultimately, understanding the dynamics of the system, including how mass changes, is crucial for accurately calculating the work done.
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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= \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?

Homework Equations


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



The Attempt at a Solution

 
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Use the fact that:

<br /> a(t) = \frac{dv}{dt}<br />

to find the velocity - time equation:

<br /> v(t) = \int{a(t) \, dt} + C_{1}<br />

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:

<br /> v(t) = \frac{dx}{dt} \Rightarrow dx = v(t) \, dt<br />

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

<br /> a(t) = \frac{dv}{dt}<br />

to find the velocity - time equation:

<br /> v(t) = \int{a(t) \, dt} + C_{1}<br />

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:

<br /> v(t) = \frac{dx}{dt} \Rightarrow dx = v(t) \, dt<br />

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

<br /> dx = \dot{x}(t) \, dt<br />

and you don't need additional information.
 
Dickfore said:
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:

<br /> dx = \dot{x}(t) \, dt<br />

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 d(\ldots) 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)
 
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