Suppose an Olympic diver who weighs 52.0 kg executes a straight dive from a 10m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water.
What is the potential energy of the diver at the apex of the dive, relative to the surface of the water?
Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed in m/s will the diver enter the water?
Does the diver do work on entering the water?
Please help me figure this out, I know that PE = mgh, but I don't seem to be getting the right answer
What is the potential energy of the diver at the apex of the dive, relative to the surface of the water?
Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed in m/s will the diver enter the water?
Does the diver do work on entering the water?
Please help me figure this out, I know that PE = mgh, but I don't seem to be getting the right answer
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PE = mgh = 52 x 9.8 x 10.8 = 5503.68 J
By the law of energy conservation:-
=>KE(surface) = PE
=>1/2mv^2 = mgh
=>v = √2gh
=>v = √[2 x 9.8 x 10.8]
=>v = 14.55 m/s
Yes some of his KE will be used against the surface tension and viscosity of water.
By the law of energy conservation:-
=>KE(surface) = PE
=>1/2mv^2 = mgh
=>v = √2gh
=>v = √[2 x 9.8 x 10.8]
=>v = 14.55 m/s
Yes some of his KE will be used against the surface tension and viscosity of water.
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PE=5503.68 J
KE=PE
1/2mv^2=mgy
mass cancels
v=√2gy
v=14.5 m/s
W=Fx
W=mgy
W=PE
maybe?
KE=PE
1/2mv^2=mgy
mass cancels
v=√2gy
v=14.5 m/s
W=Fx
W=mgy
W=PE
maybe?