Calculating Work Done by a Variable Force

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To calculate the work done by a variable force F=bx^3, where b is 3.7 N/m^3, one must recognize that the force changes with position, making the simple multiplication of force and distance insufficient. Instead, the correct approach involves using integration to account for the variable nature of the force over the distance from x=0.0m to x=2.2m. The definite integral of the force function from 0 to 2.2m yields the total work done. The final calculation results in 21.6 J, which rounds to 22 J. Understanding how to apply integration in this context is crucial for solving similar problems.
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Homework Statement



A force F=bx^3 acts in the x-direction. How much work is done by this force in moving an object from x=0.0m to x=2.2m? The value of bis 3.7 N/m^3


Homework Equations



Work=Force*Distance


The Attempt at a Solution



Ok since the above equation is true, then the problem would solution would be obtained by simply multiplying our force which is (3.7N/m^3) by the given distance which would be 2.2m. The thing is that once you do that, the problem is over and there is still a variable there bothering us which is x^3. What am I supposed to do with that variable?
 
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(3.7N/m^3)

This is not the force, this is a constant. The force is b*x^3. Since it contains the x^3 term, the force is variable and you can't use W=Fd. Do you know integration?
 
yes, but how would i use it in this case?edit: ok i figured it out, it should be the definite integral from 0-2m of F right?
 
Last edited:
ecthelion4 said:
yes, but how would i use it in this case?


edit: ok i figured it out, it should be the definite integral from 0-2m of F right?

Yes, but the upper limit is 2.2m, right? :smile:
 
that was exactly it. The answer was 21.6 which rounded up to 22J . Thanks you were a great help!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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