A small block of mass m = 125 g is released from rest at point A along the horiz
ID: 2251932 • Letter: A
Question
A small block of mass m = 125 g is released from rest at point A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 32.0 cm as shown in the figure below.
(a) Calculate the gravitational potential energy of the block-Earth system when the block is at point A relative to point B
(b) Calculate the kinetic energy of the block at point B.
(c) Calculate its speed at point B.
(d) Calculate its kinetic energy when the block is at point C.
(e) Calculate the potential energy when the block is at point C.
Explanation / Answer
point A is on the rim of the bowl.
Therefore it is a vertical distance of 0.32m from point B.
Therefore relative to point B PE = mgy = .125kg*9.80m/s^2*0.32m = 0.392J
. Now when the ball moves to point B all of the PE is converted to KE so KB = 0.392J.
If KE = 0.392 = 1/2*m*v^2, we get v= sqrt(2*KE/m) =sqrt(2*0.392/.125) = 2.5m/s.
Now when the ball moves up to point C, we calculate its PE = mgy = .125*9.8*2/3*0.32 = 0.261 J.
Since Total energy is conserved then KE = 0.392J - 0.261J = 0.131J