Work Performed by Force F on Body of Mass m up a Hill

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The problem involves a body of mass m being slowly pulled up a hill by a force F, which is directed along the trajectory's tangent. The key point is that the term "slowly hauled" indicates that the body moves at a constant speed, implying no acceleration. This means that the applied force F must equal the sum of opposing forces, including friction and gravitational components. The work done by force F can be calculated using the height of the hill h, the base length l, and the coefficient of friction K. Understanding these dynamics is crucial for solving the problem accurately.
danilo_rj
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I've got a problem here from Irodov (it is a very well known physics book).
1.121) A body of mass m was slowly hauled up the hill by a force F which at each pont was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and the coefficient of friction K.

There is picture, but is not necessary 'cause the text of the problem is enough.
Anyway, I didn't understand why the body has no acceleration when it is being hauled by the force F. And why it is said that the body was slowly hauled?
 
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Presumably, F is just the applied force needed to overcome the other forces acting on the body, not the net force. Assume that "slowly hauled" means no acceleration.
 
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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