Answer:
Given:
mass = 100 g = 0.1 kg
g = 10 m/s^{2}s
2
height of A above O = 5 cm = 0.05 m
To find :
a) Potential energy at 'A' and where does it come from.
b) Total energy at 'A'.
c) Kinetic energy at mean position.
d) Kinetic energy and potential energy of the bob at position whose height above is 2 cm.
Solution:
(a) The value of potential energy of the bob at 'A' = m * g * h
= 0.1 * 10 * 0.05 = 0.05 J
It is stored due to the work done on it for raising .
(b). The value of total energy of the bob at position 'A' = P.E. + K.E.
= 0.05 + 0 = 0.05 J
(c). At mean position , total P.E. change to K.E.
So , kinetic energy of the Bob at mean position = 0.05 J
(d). P.E. at 2 m height = m * g * h
= 0.1 * 10 * 0.02 = 0.02 J
K.E. at 2 m height = total energy - P.E. at 2 m height
= 0.05 - 0.02
= 0.03 J
Potential energy at 'A' is 0.05 J and it is stored due to the work done on it .
Total energy at 'A' is 0.05 J .
Kinetic energy at mean position is 0.05 J .
Kinetic energy is 0.03 J and potential energy of the bob is 0.02 J at position whose height above is 2 cm.