Calculating work

indietro

1. The problem statement, all variables and given/known data
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^{[tex]\circ[/tex]}N] and T2 is 1295N [E45^{[tex]\circ[/tex]}N]


2. Relevant equations
so the work by gravity is 12500 J
the work by T1 is 4575 J (opposite direction to work by gravity)

3. 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[tex]\theta[/tex][tex]\Delta[/tex]r
= 1295*cos45*(5)
= 4579 J

???? - thnx!

fatra2

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

indietro

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