In the figure below, a 2.10 g ice flake is released from the edge of a hemispher
ID: 1438893 • 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 25.0 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 (e) If the mass of the flake were doubled, would the magnitudes of the answers to (a) through (d) increase, decrease, or remain the same? increase decrease remain the sameExplanation / Answer
a) Work done on flake by weight (gravity) = m*g*h
=0.0021*9.81*0.25 = 5.15x10^-3 J
b) change in potential energy of flake-Earth system = 0
c) if PE = 0 at bottom then PE at top = mgh = 5.15x10^-3 J
d) If PE = 0 at top the PE at bottom = -mgh = - 5.15x10^-3 J
e) All of these answers are linear in m. If you double the mass, the results double. so increases.