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

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SUMMARY

The speed of a 1.3 kg block dropped from a height of 0.41 m, tied to a string around a uniform disk pulley with a mass of 0.30 kg and a radius of 7.2 cm, can be calculated using kinematic equations. The initial calculation of the final speed (Vf) as 2.836 m/s using the formula Vf = sqrt(2 * g * h) is incorrect because it neglects the effect of the pulley on the block's acceleration. The presence of the pulley alters the dynamics, requiring the application of rotational motion equations to accurately determine the block's speed.

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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?
 

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