Tbr cbt 3 #49

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AFLATPEG

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TBR CBT 3 #49


For the pulley system shown below, how much energy is needed to lift a box weighing 4 kg to a height of 5 meters? (Note: cos 60˚ = 0.50, sin 60˚ = 0.87.)

C.) is the answer.

why can't i use Work = Fd cos(theta) to solve this? read the solution & understand it. but i ask because that is the equation that first popped up in my head so i went with it, but ended up with a wrong answer.

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The amount of worked needed to lift the box is mg times the distance. Pulley can change the force that you need to apply and trade it for the distance but the total work stays the same (ignoring any friction related complications).
 
The amount of worked needed to lift the box is mg times the distance. Pulley can change the force that you need to apply and trade it for the distance but the total work stays the same (ignoring any friction related complications).


but isn't Word= Fdcos(theta)

theta being the angle BETWEEN the direction its moving & the force that is applied?

isn't the force making the object move the force that is pulling on the string pulley and the distance is object going up vertically. so the angle between those two would be theta?


doesn't.. F=pulling force and d=5m up. so theta would be 90-60deg= 30 deg.?

just curious why.

basically trying to be analogous to this photo below:

phpAldNHG.png
 
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The force moving up the block is the tension force from the hook on the pulley pulling it up, it has nothing to do with the angle at which the rope is being pulled.

I would not even worry about forces anyway, since whoever posted that the force cannot be equal to the the weight of the block is correct - you do need to accelerate and decelerate the block. It's much easier to approach this from energetic point of view - W=ΔU. The potential energy increases by mgd, so the work done on the block is mgd.
 
The force moving up the block is the tension force from the hook on the pulley pulling it up, it has nothing to do with the angle at which the rope is being pulled.

I would not even worry about forces anyway, since whoever posted that the force cannot be equal to the the weight of the block is correct - you do need to accelerate and decelerate the block. It's much easier to approach this from energetic point of view - W=ΔU. The potential energy increases by mgd, so the work done on the block is mgd.

energetics! yes, always easier. thank you :)
 
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