The Work - Energy Theorem in a pulley?

AI Thread Summary
The discussion focuses on applying the work-energy theorem to a pulley system with a frictional block. The key points include calculating the potential energy lost by the descending block and the corresponding kinetic energy gained by the other block. The coefficient of kinetic friction is specified, which affects the work done against friction. Participants are encouraged to quantify the changes in potential energy, kinetic energy, and the work done to solve for the speed of the 6.00-kg block after descending 1.50 m. The conversation emphasizes understanding the balance of energy transformations in the system.
erik-the-red
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Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the tabletop is \mu_k=0.250. The blocks are released from rest.

Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

\Delta * K = (1/2)(m)(v_f^2 - v_i^2)

But, it was at rest, so (1/2)(m)(v_f^2) is what matters.

What do I do next?
 

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You will have a loss in PE balanced by a gain in KE and work done.

What is it that loses PE, and by how much ?

What are the gains in KE and by how much ?

What work is done and how much ?
 
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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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