Posted by Sandhya on Monday, October 27, 2008 at 11:30pm.
A 1.00 kg hollow ball with a radius of 2.2^10^1 m, filled with air, is released from rest at the bottom of a 1.50 m deep pool of water. How high above the water does the ball shoot upward if the upward force exerted by the fluid on the ball 40 N? Neglect all frictional effects, and neglect the ball's motion when it is only partially submerged.
I used energy consideration & got the following answer. But the answer is incorrect.Please point out where i have gone wrong
Under the water you have the net force on the ball; F= 40 N. That force acts over
a distance D (1.5 m in this case) and thus does work on the ball giving it some
KE as it leaves the water: KE = FD. = 40*1.5 = 60 J
To find out how high the ball rises in the air, set that initial KE to the PE at the top
of the motion: KE = FD = 60 J = mgH. Now solve for H. where m = 1 kg
H = 60/1*9.8
= 6.12 m

physics  drwls, Tuesday, October 28, 2008 at 7:39am
Perhaps it has something to do with the radius of the ball. When the center of the ball has risen 1.50  0.44 m, it is no longer fully submerged. They want you to assume that work stops then. It is not clear where to measure the potential energy change from: the surface or one radius below.
The answer you get depends upon the approximations made
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