A pendulum bob of mass 0.5 kg is released from rest from a horizontal position. The length of the pendulum is 2 m . When the bob arrives at the lowest point of its path, the tension in the string is measured to be 13 N . The percentage of the initial potential energy that was dissipated against air resistance during the fall is: [Use g = 10 m s ⁻² ]

Step-by-step solution

Let the speed of the bob at the lowest point be v . At the lowest point, the net force towards the center provides the necessary centripetal acceleration: T - mg = mv^2 L Substitute the given values ( m = 0.5 kg , L = 2 m , T = 13 N , g = 10 m s ⁻² ): 13 - (0.5 10) = 0.5 v^2 2 13 - 5 = 0.25 v^2 8 = 0.25 v^2 v^2 = 32 m ^2 s ⁻² The final kinetic energy of the bob at the lowest point is: K = 1 2 mv^2 = 1 2 0.5 32 = 8 J The initial potential energy of the bob at the horizontal position (taking the lowest point as the r