Work and Kinetic Energy of barrel

AI Thread Summary
To calculate the work done against gravity when raising a 45.0 kg barrel by 6.60 m, the formula W = mgh yields 2910 J, which represents the gravitational work. The net work done on the barrel is determined by considering both the work done by the cable and the work done against gravity. The cable exerts a force of 710 N, which contributes positively to the net work. By applying the equations for both gravitational work and cable work, the total net work can be calculated. Understanding the distinction between work done against gravity and net work is crucial for solving the problem accurately.
ScienceGirl90
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Homework Statement


A 45.0 kg barrel is raised 6.60 m (from rest) by cable that exerts 710 N on the barrel. Neglecting any frictional energy losses, how much work is done against gravity to lift the barrel and what is the net work done on the barrel?



Homework Equations


W=m*g*h
W=F*D


The Attempt at a Solution


I am confused about the wording, "against gravity". Also when I tried to find the net work done on the barrel I did W=(45.0kg)(9.80m/s/s)(6.60m)= 2910 J. Is that right?
 
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against gravity means you are lifting the object, opposing the action of gravity.
So your W(net) = W(grav) + W(cable)

now apply W=mgh for both cases, (keeping in mind the signs), and add them to get your net work.
 
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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