In the figure below, a 2.10 g ice flake is released from the edge of a hemispher
ID: 1469709 • Letter: I
Question
In the figure below, a 2.10 g ice flake is released from the edge of a hemispherical bowl whose radius r is 17.0 cm. The flake–bowl contact is frictionless.
(a) How much work is done on the flake by the gravitational force during the flake's descent to the bottom of the bowl?
J
(b) What is the change in the potential energy of the flake–Earth system during that descent?
J
(c) If that potential energy is taken to be zero at the bottom of the bowl, what is its value when the flake is released?
J
(d) If, instead, the potential energy is taken to be zero at the release point, what is its value when the flake reaches the bottom of the bowl?
J
Explanation / Answer
Mass of ice flake= 2.1 g = 0.0021 kg
height from bottom = radius= 17 cm= 0.17 m
a) Work done on flake by weight (gravity) = m*g*h
=0.0021*9.81*0.17 = 3.5 x10^-3 J
b) change in potential energy of flake-Earth system = 0
c) if PE = 0 at bottom, then the PE at top = mgh = 3.5x10^-3 J
d) If PE = 0 at top, then the PE at bottom = -mgh = -3.5x10^-3 J