In the figure, a 2.00 g ice flake is released from the edge of a hemispherical b

Question

In the figure, a 2.00 g ice flake is released from the edge of a hemispherical bowl whose radius r is 22.0 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? (b) What is the change in the potential energy of the flake-Earth system during that descent? (c) What is the speed of the flake when it reaches the bottom of the bowl? (d) If we substituted a second flake with twice the mass, what would its speed be?

Explanation / Answer

If we take gravitational potential to be zero at bottom of bowl the

a) work done on flake by gravitational field = increase in gravitational potential energy = 0 - mgr = - 0.002*9.81*0.22

W = -4.32*10-3 joules

b) Change in potential energy of flake-earth system is same as change in work done by gravitational field which is

4.32*10-3 joules

c) KE of flake at bottom of bowl = decrease in potential energy =4.32*10-3

1/2*m*V2 = 4.32*10-3

V2 = 4.32

Vbottom = 2.08 m/s

c) if m'= 2m = 0.004 kg

then

1/2*2m*V2 = 4.32*10-3

V2 = 2.16

Vbottom = 1.47 m/s

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