A pendulum oflength 2m lift at p.When it reaches q
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Samaritan
The answer seems to be 6m/s.
This is a problem in physics that potential energy of the pendulum is converted into kinetic energyand is dissipated into the air.
Therefore, PE = KE = LOSS OF ENERGY………………………………….(1).
According to the problem, energy loss is 10 % of PE=0.10PE.
So, equation 1 becomes,
PE + KE+0.1 PE=0.9PE =KE.
THEREFORE,
0.9 Mgh =0.5 V2. V2 = 1.8 gh. V= speed of the lowest point.
h= length of the pendulum= 2 m, g = 9.81 m/sec2.
putting the values of g and h.
Velocity at B, V = 5.94 m/sec
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6m/sec is the answer
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