Calculating Work: Piano Lowering 5m, 2500N Gravity, T1 1830N, T2 1295N

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SUMMARY

The discussion focuses on calculating the work done by forces acting on a piano being lowered 5 meters, with a gravitational force of 2500 N and tensions T1 at 1830 N [W60°N] and T2 at 1295 N [E45°N]. The work done by gravity is confirmed to be 12500 J, while the work done by T1 is calculated as 4575 J. The participant initially miscalculated the work done by T2, arriving at approximately 4000 J instead of the correct value of 4579 J, leading to a net work of zero when all forces are considered.

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Homework Statement


There is a piano that is being lowered 5 m. It has gravity force acting in y-direction (down) of 2500 N. Then it has two ropes attached, they are vectors that are pointing up and out. so T1 is 1830N [W60\circN] and T2 is 1295N [E45\circN]


Homework Equations


so the work by gravity is 12500 J
the work by T1 is 4575 J (opposite direction to work by gravity)

The Attempt at a Solution


so it would make sense that the work by T2 is somewhere close to 7000 J so that the net work is 0. but when i calculate it i get around 4000 J. Can someone check my math...

For T2:
W = Fcos\theta\Deltar
= 1295*cos45*(5)
= 4579 J

? - thnx!
 
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Hi there,

I believe you have a problem with the angle of T1. Becausem, when I verified your calculations, I come to precisely 0 net work.

Cheers
 
oh yes thank you! i was using an angle that gave the component perpendicular to motion :)
 

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