The student says that a ball dropped from a height of 3.0 meters would have less than 96 joules of kinetic energy when it hits the ground. Is she correct and why?
Type Answer HERE:
No, the student is not correct. The kinetic energy of an object can be determined using the equation KE = 1/2mv^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object. The height from which the ball is dropped does not directly determine its kinetic energy. However, it can indirectly affect the kinetic energy if it affects the velocity of the ball. In this case, assuming no air resistance, the velocity of the ball when it hits the ground can be determined using the equation v = sqrt(2gh), where g is the acceleration due to gravity and h is the initial height. Substituting the given values, v = sqrt(2*9.8*3) = 7.67 m/s. Assuming the mass of the ball to be 1 kg, the kinetic energy can be calculated as KE = 1/2 * 1 * (7.67)^2 = 29.4 J. Therefore, the ball would have less than 96 joules of kinetic energy when it hits the ground.
To determine whether the student is correct or not, we need to calculate the kinetic energy of the ball when it hits the ground.
The formula for kinetic energy is given by:
KE = 0.5 * m * v^2
Where KE is the kinetic energy, m is the mass of the ball, and v is the velocity of the ball.
Given that the ball is dropped from a height of 3.0 meters, we can use the formula for potential energy to find the velocity of the ball when it hits the ground:
PE = m * g * h
Where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the ball is dropped.
Since the ball is initially at rest, its potential energy is equal to its initial kinetic energy:
PE = KE
Substituting the values, we have:
m * g * h = 0.5 * m * v^2
Canceling out the mass, we get:
g * h = 0.5 * v^2
Rearranging the equation:
v^2 = 2 * g * h
Substituting the values of g (9.8 m/s^2) and h (3.0 m), we can calculate the velocity of the ball when it hits the ground:
v^2 = 2 * 9.8 m/s^2 * 3.0 m
v^2 = 58.8 m^2/s^2
v ≈ 7.67 m/s
Now that we know the velocity, we can calculate the kinetic energy using the formula:
KE = 0.5 * m * v^2
Since we don't have the mass of the ball, we can assume a mass of 1 kg for simplicity.
KE = 0.5 * 1 kg * (7.67 m/s)^2
KE ≈ 29.57 joules
Since the calculated kinetic energy is less than 96 joules, the student is correct. The ball dropped from a height of 3.0 meters would have less than 96 joules of kinetic energy when it hits the ground.