Question

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 when the ball is at its lowest point (θ = 0).
a) What is the gravitational potential energy, in joules, of the ball before it is released?
b) What will be the speed of the ball, in meters per second, when it reaches the bottom?

Answer 1

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:  

$K.E=\frac{I}{2}*mv^{2}$

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:

$0.2275=\frac{1}{2} *0.25*v^{2}$

$v^{2}=\frac{2*0.2275}{0.25}$

$v^{2}=1.82$

$v=\sqrt{1.82}=1.3491$ ≈ $1.35m/s$

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

 

Answer 2

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$