this is a pendulum problem?
a 0.157 kg simple pendulum is pulled aside from rest at a height of 0.252 m above its lowest point. find its speed when it
a.) is 0.149 m above the low point.
b.) reaches height of 0.252 m on the opposite side of its swing
c.) is at the low point of its swing.
Comments
it's all about energy
potential and kinetic.
we will start with the initial potential energy at a height of .252 m
potential energy = m*g*h = 0.157*9.81*.252 = 0.388123 kg m^2/s^2
A) it converts some of it's potential energy to kinetic energy
0.5*m*V^2 = m*g*(.252-0.149)
v^2 = 2.021 m^2/s^2
v=1.42157 m/s
B)it will come to rest at a distance equal to the original drop height. All of the potential energy get converted to kinetic at the bottom of the swing and then gets converted back to potential at the same height on the other side.
c) 0.5*m*V^2 = m*g*0.252
V^2 = 4.944 m^2/s^2
V = 2.2236 m/s