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 lower most point B, given that it dissipated 5% of its initial energy against air resistance?

• A. 5 m/s
• B. 5.5 m/s
• C. 5.3 m/s
• D. 4.4 m/s

Hint:

When energy is dissipated due to air resistance, use the fraction of energy remaining to calculate the final speed.

Answer 1

The Correct Option is C

The potential energy of the pendulum at the top (point A) is converted into kinetic energy at the bottom (point B), with some energy lost due to air resistance.
The initial potential energy is given by: \[ E_{\text{initial}} = mgh = mgL(1 - \cos \theta) \] where \( L = 1.5 \, \text{m} \) is the length of the pendulum, and \( h \) is the vertical height. At the lowest point, the kinetic energy is: \[ E_{\text{kinetic}} = \frac{1}{2}mv^2 \] The energy dissipated due to air resistance is 5% of the total energy, so the remaining energy is 95%.
Therefore: \[ \frac{1}{2} mv^2 = 0.95 \times mgh \] Substituting \( h = L \), we get: \[ \frac{1}{2} mv^2 = 0.95 \times mgL \] Simplifying: \[ v^2 = 2 \times 0.95 \times gL \] \[ v = \sqrt{2 \times 0.95 \times 9.8 \times 1.5} \] \[ v \approx 5.3 \, \text{m/s} \]
Thus, the correct answer is (c).