Question

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

a) 4.9 m/s

b) 5.5 m/s

c) 4.7 m/s

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Answer 1

$\huge\bold{\textbf{\textsf{{\color{cyan}{Answer}}}}} = 5.34 \frac{m}{s}$

Given

1 = 1.5 m

energy lost = 5%

Solution

a) In horizontal position

P.E. = mgh = mgl K.E. = 1/2 mv^2 = 0

Total energy E1 = mgl

But 5% energy is dissipitaed. Thus E1' = 95% E1

b) At lowermost position

P.E. = mgh = 0 K.E. = 1/2 mv^2 Total energy E2 = 1/2 mv^2

By law of conservation of energy

E1' = E2

95% mgl = 1/2 mv^2 v^2 = 95/100 × 10 × 1.5 × 2

v^2 = 28.5

v = 5.34 m/s

Thus, speed of the bob at lowermost position will be 5.34 m/s.