Answer:
a. 0.23J
b. 1.35 m/s
Explanation:
a. U = mgh where where m = mass of the object, g = acceleration due to gravity, and h = height
h = L - Lcos(θ) where L = length of the rope, and θ = angle with respect to vertical.
Therefore, U = mg(L - Lcos(θ))
U = 0.25 * 9.8 (0.65m - 0.65cos(31° ))
U = 0.2275 ≈ 0.23J
The gravitational potential energy of the ball before it is released = 0.23J
b. To determine the velocity of the object at the bottom of its motion, all of the energy has gone from gravitational potential into kinetic since at the bottom, the problem says that U = 0. The kinetic energy of an object is given by the following equation:

where m = mass of the object and v = velocity of the object. Since we know that all of the energy was transferred into kinetic energy at the bottom, we can conclude that:



≈ 
Therefore, the speed of the ball when it reaches the bottom = 1.35m/s