First slide

Numericals based on work energy theorem

Question

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

Moderate

A

5.1 m/sec
B

5.5 m/sec
C

5.3 m/sec
D

4.4 m/sec

Solution

At the point A the energy of the pendulum is entirely potential energy. At point B energy of the pendulum is entirely kinetic energy. It means that as the bob pendulum lowers from A to B, the potential energy is converted in to kinetic energy. Thus at B, K.E=P.E. But 5% of potential energy is
dissipated against air resistance.
$\begin{array}{r} \frac{1}{2} m v^{2} = \frac{95}{100} m g h \\ v^{2} = \sqrt{\frac{19 \times 9.8 \times 1.5}{10}} \end{array} v = 5.3 m / s$

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