Is my calculation for work done correct?

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The calculation for work done by a uniformly changing force was initially incorrect, as the method used did not account for the average force. To find the correct work done, one can either use calculus to derive the force vs. displacement equation, graph the force-displacement to find the area under the curve, or apply the average force approach. The correct formula for work done in this scenario involves calculating the average force over the displacement. After reevaluation using the suggested methods, the final calculation was confirmed to be accurate.
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Homework Statement


Force magnitude at the beginning of the road s = 12 m is F1 = 10 N , at the end of the road F2 = 46 N. The force is acting in object's displacement direction. Calculate work done by uniformly changing force.

Homework Equations


A = Fs

The Attempt at a Solution


A = Fs = (F2-F1)s = 432 J

Is it correct?
 
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kaspis245 said:

Homework Statement


Force magnitude at the beginning of the road s = 12 m is F1 = 10 N , at the end of the road F2 = 46 N. The force is acting in object's displacement direction. Calculate work done by uniformly changing force.

Homework Equations


A = Fs

The Attempt at a Solution


A = Fs = (F2-F1)s = 432 J

Is it correct?
No.

You can write the force vs. displacement equation and do the calculus for the work done, or make a graph of the force-displacement and find the area under it, or , more easily since the force is uniformly changing with distance, try the average force approach when using your formula. Use all three methods if you want and compare results. Your approach is not correct.
 
What about now:

b4td1v.jpg
 
kaspis245 said:
What about now:
Looks Good!

b4td1v.jpg
 
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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