Question

A 1.57 g ice flake is released from the edge of a hemispherical bowl whose radius r is 26.9 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? (b) What is the change in the potential energy of the flake-Earth system during that descent? (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? (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?

Answer 1

Explained

Step-by-step explanation:

work is done on the flake by the gravitational force during the flake's descent to the bottom of the bowl

w= mg×r= 1.57×10^{-3}×9.8×0.269 = 0.004138 j

= 4.13×10^{-3} J

b) ΔU = -4.13×10^{-3} 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

U= mgr = 4.13×10^{-3} J

d) U= -mgr

= -4.13×10^{-3} J