In the figure above, the pulley has negligible mass, and both it and the incline
ID: 1351636 • Letter: I
Question
In the figure above, the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.0 kg, and block B has a mass of 2.0 kg, and angle ? is 30°. If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block B has fallen 30 cm? In the figure below, the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.0 kg, and block B has a mass of 2.0 kg, and angle ? is 30°. If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block B has fallen 30 cm?
Explanation / Answer
Potential energy loss of block B = mb*g*h
If block B moves h down, block A moves h up along the plane. The change in elevation of block A is then h*sin and it gains potential energy of ma*g*h*sin.
The net loss of potential energy is then
PE(loss) = mb*g*h - ma*g*h*sin
here h = 0.30 m
ma = 1.0 kg
mb = 2.0 kg
= 30º
and here kinetic energy is equal to loss of potential energy so
KE = PE(loss)
KE = 2*9.8*0.3 - 1*9.8*0.3*(1/2) = (3/2)*9.8*0.3 = 4.41 J