What is the speed of a block dropped from 0.41m with a pulley involved?

AI Thread Summary
The discussion revolves around calculating the speed of a 1.3 kg block dropped from a height of 0.41 m, considering its connection to a pulley with a mass of 0.30 kg and a radius of 7.2 cm. The initial calculation using kinematics yielded a speed of 2.836 m/s, assuming constant acceleration due to gravity. However, participants noted that the presence of the pulley affects the block's acceleration, suggesting that the acceleration is not simply 9.81 m/s². Additionally, the impact of attaching a lead weight to the pulley on the block's speed was questioned, prompting a discussion on how it would alter the system's dynamics. Understanding the interaction between the block and the pulley is crucial for accurate speed calculations.
MellowOne
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Homework Statement


A 1.3 kg block is tied to a string that is wrapped around the rim of a pulley of radius 7.2 cm. The block is released from rest.

(a) Assuming the pulley is a uniform disk with a mass of 0.30 kg, find the speed of the block after it has fallen through a height of 0.41 m.

(b) If a small lead weight is attached near the rim of the pulley and this experiment is repeated, will the speed of the block increase, decrease, or stay the same? Explain.

Homework Equations


Kinematics?

The Attempt at a Solution


Since the block was just falling I assumed you could just use the 3rd kinematics and solve for vf with vi = 0. Substituting the variables I got Vf = sqrt (2 x 9.81m/s^2 x .41m) and the answer came out to be 2.836m/s. Any reason why this wouldn't work?
 
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You can't assume a=9.81 m/s^2, though. If the block is attached to the pulley by a string, don't you think that might change its 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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