A 0.200 kg ball sits on a shelf 2.00 meters from the floor and then falls to a point 1.00 meter from the floor ignoring any affects due to fiction at this position_______

To solve this problem, we can use the principles of conservation of mechanical energy.

The initial energy of the ball on the shelf is potential energy, given by the formula:
PE_initial = m * g * h_initial

Where:
m = mass of the ball = 0.200 kg
g = acceleration due to gravity = 9.8 m/s^2
h_initial = initial height = 2.00 meters

Substituting the given values, we get:
PE_initial = 0.200 kg * 9.8 m/s^2 * 2.00 meters = 3.92 J

As the ball falls, its potential energy is converted into kinetic energy. At the point 1.00 meter from the floor, all of the potential energy is converted to kinetic energy, as there is no friction mentioned.

The kinetic energy of the ball is given by the formula:
KE = 0.5 * m * v^2

Where:
m = mass of the ball = 0.200 kg
v = velocity of the ball

Since the ball falls freely due to gravity, we can use the equation for calculating the final velocity of an object in free fall:
v_final^2 = v_initial^2 + 2 * a * d

Where:
v_initial = initial velocity = 0 m/s (since the ball was at rest on the shelf)
a = acceleration due to gravity = 9.8 m/s^2 (downwards)
d = distance traveled = h_initial - h_final = 2.00 meters - 1.00 meter = 1.00 meter (downwards)

Substituting the given values, we get:
v_final^2 = 0^2 + 2 * 9.8 m/s^2 * 1.00 meter
v_final = sqrt(19.6) m/s = 4.43 m/s (rounded to two decimal places)

Now, we can calculate the kinetic energy using the given formula:
KE = 0.5 * 0.200 kg * (4.43 m/s)^2
KE = 0.5 * 0.200 kg * 19.61 m^2/s^2 = 1.961 J (rounded to three decimal places)

Therefore, the kinetic energy of the ball at the point 1.00 meter from the floor is 1.961 Joules.

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