In the figure below, a 2.90 g ice flake is released from the edge of a hemispher

Question

In the figure below, a 2.90 g ice flake is released from the edge of a hemispherical bowl whose radius r is 16.0 cm. The flake-bowl contact is frictionless. What is the speed of the flake when it reaches the bottom of the bowl? If we substituted a second flake with twice the mass, what would its speed be? A worker pushed a 27 kg block 9.0 m along a level floor at constant speed with a force directed 33 degree below the horizontal, (a) If the coefficient of kinetic friction is 0.18, how much work was done by the worker's force? What was the increase in thermal energy of the block-floor system?

Explanation / Answer

use conservation of energy
potential energy at top = kinetic energy at bottom
m*g*h = 0.5*m*v^2
g*h = 0.5*v^2
v = sqrt (2*g*h)

a)
v = sqrt (2*g*h)
= sqrt (2*9.8*0.16) {here h = r = 16 cm = 0.16 m}
=1.77 m/s
Answer: 1.77 m/s

b)
v = sqrt (2*g*h)
speed is independent of mass
so, speed will be same as in previous case
Answer: 1.77 m/s

Only 1 question at a time please. Thanks

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