The length of a pendelum is 25cm, and the mass of teh bob is 300g. It is displac

Question

The length of a pendelum is 25cm, and the mass of teh bob is 300g. It is displaced from the equilibrium position and released. At the lowest point of the trajectory the KE of the pendelum is 0.265J.

A). how fast is the pendulum moving when at the lowest point?

B). from wht height ( above the lowest point of the trajectory) was it released?

C). what is the total mechanical energy of the pendulum?

D)/ The pendulum is passing through a point 5 cm above its equilibrium position ( its lowest point)

----FIND ITS PE

------Find its KE

Explanation / Answer

a) The kinetic energy at the lowest point is
KE = 0.265 J
(1/2) m v2 = 0.265 J
v2 = 2 x 0.265 / 0.3 kg
v2 = 1.767
v = 1.33 m/s
b) The total mechanical energy is conserved in this periodic motion
So the kinetic energy at the lowest point is equal to the potential energy at the highest point
m g h = (1/2) m v2
The kinetic energy at the lowest point is 0.265 J
h = 0.265 J / m g
h = 0.265 / 0.3 kg x 9.81 m/s2
h = 0.09 m = 9 cm
c) The total mechanical energy of the pendulum
E = KE + PE
At the lowest position, the mechanical energy is equal to the kinetic energy, at that position, the potential energy is zero
E = KE
E = 0.265 J
d) The potential energy at the point is
PE = m g h
PE = 0.3 kg x 9.81 m/s2 x 0.05 m
PE = 0.147 J
Considering the total mechanical energy
E = KE + PE
KE = E - PE
KE = 0.265 J - 0.147 J
KE = 0.118 J

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