a 3 X 10^2 kg piano is being lifted at a steady speed from ground level straight up to an apartment 10.0 m above the ground. the crane that is doing the lifting produces a steady power of 4.00 X 10^2 W. how much time does it take to lift the piano and how do you mathmatically write this out?
2007-03-06 14:37:03 · 4 answers · asked by tico 1 in Science & Mathematics ➔ Physics
Power = Work / time
In this case Work = mgH
So,
P = mgH / t
or
t = mgH/P
where m = 300kg, g=9.8m/s/s, H=10m, and P = 400 W
t = 73.5 seconds
2007-03-06 14:43:21 · answer #1 · answered by Dennis H 4 · 0⤊ 0⤋
Total energy requirement = power x time
mgh = P x t
3x10^2x9.8x10 = 4.00x10^2 x t; all quantities in MKS so t is in seconds,
t = (3x98)/4 = 73.5 s = 1 m 13.5 s.
It will take some what more time in actual practice as we have not accounted for any losses due to frictional forces and also piano is not brought up all the time remaining at rest. So there is some average kinetic energy of the piano as well as the crane parts.
2007-03-06 22:54:10 · answer #2 · answered by Let'slearntothink 7 · 0⤊ 0⤋
The potential energy of the pinao at its full height is
Ep = mgh m=mass, g=gravity, h=height
Ep = 10*9.8*300 = 29,400 joules
Power = 400 watts. 1 watt = 1 joule/second
= 400 J/s
If Ep is in joules and P is in joules/second then to work out the number of seconds divide Ep/P - the units cancel out leaving only seconds.
time = Ep/P = 29,400/400 = 73.5 seconds
2007-03-06 22:50:59 · answer #3 · answered by Possum 4 · 0⤊ 0⤋
work is defined as distance x force. Power is work per unit of time. Force is measured in Newtons. 1 Newton = 1 kg m/sec^2. Unless I'm mistaken, the time it takes can be calculated by work/power.
2007-03-06 23:03:02 · answer #4 · answered by misoma5 7 · 0⤊ 0⤋