A simple pendulum,2.6 m in length,is released with a push when the support string is at an angle of 45.0° from the vertical.If the initial speed of the suspended mass is 1.25 m/s when at the release point,what is its speed at the bottom of the swing? (g = 9.8 m/s²) A)4.1 m/s. B)3.9 m/s C)4.1 m/s D)16 m/s E)7.2 m/s

Question

Quizplus · Accepted Answer

Using conservation of energy, we can relate the potential energy at the release point to the kinetic energy at the bottom of the swing:
mgh = (1/2)mv^2
where m is the mass of the suspended object, g is the acceleration due to gravity, h is the height above the bottom of the swing, and v is the speed at the bottom of the swing.
At the release point, the height above the bottom of the swing is h = L(1-cos(theta)) = 2.6(1-cos(45°)) = 1.68 m. Plugging in the given values, we get:
mgh = (1/2)mv^2
(4.4 kg)(9.8 m/s^2)(1.68 m) = (1/2)(4.4 kg)(1.25 m/s)^2 + (4.4 kg)v^2
Solving for v, we get:
v = sqrt[(2/7)(9.8)(1.68 - 0.78)] = 4.1 m/s.
Therefore, the answer is A.

A Simple Pendulum,2