A stone is thrown from the edge of a cliff into the ocean below. It is thrown somewhat upward, but mostly outward. The stone has a mass of 2.0kg and is thrown with a speed of 10m/s from an initial height of 5.0m above the ocean.
A. What is the kinetic energy of the stone when it is thrown?
B. What is the Potential energy of the stone relative to the ocean when it is first thrown?
C. What is the kinetic energy of the stone just as it hits the water?
D. What is the speed with which the stone hits the water?
a. KE=1/2 m v^2
b. PE=mgh
c. KEfinal=initial KE + initial PE
d. 1/2 mv^2=final ke
solve for v.
To answer these questions, we need to understand the concepts of kinetic energy, potential energy, and the conservation of energy.
A. To calculate the kinetic energy of the stone when it is thrown, we use the formula:
Kinetic energy = 1/2 * mass * velocity^2
Here, the mass of the stone is given as 2.0kg, and the velocity is given as 10m/s. Plugging these values into the formula, we get:
Kinetic energy = 1/2 * 2.0kg * (10m/s)^2
Kinetic energy = 1/2 * 2.0kg * 100m^2/s^2
Kinetic energy = 100J (Joules)
Therefore, the kinetic energy of the stone when it is thrown is 100 Joules.
B. The potential energy of an object is the energy it possesses due to its position relative to other objects. In this case, the potential energy of the stone relative to the ocean when it is first thrown can be calculated using the formula:
Potential energy = mass * acceleration due to gravity * height
The mass of the stone is given as 2.0kg, the acceleration due to gravity is approximately 9.8m/s^2, and the height is given as 5.0m. Plugging these values into the formula, we get:
Potential energy = 2.0kg * 9.8m/s^2 * 5.0m
Potential energy = 98 J (Joules)
Therefore, the potential energy of the stone relative to the ocean when it is first thrown is 98 Joules.
C. The kinetic energy of the stone just as it hits the water can be calculated using the same formula as in part A. However, the velocity of the stone just as it hits the water is not given. To calculate this, we need to use the concept of conservation of energy. The total energy of the stone is conserved, meaning that the sum of its kinetic and potential energy remains constant.
Since we know the initial potential energy and the initial kinetic energy (which is 100 Joules from part A), we can write the equation:
Initial potential energy + Initial kinetic energy = Final potential energy + Final kinetic energy
98J + 100J = Final potential energy + Final kinetic energy
Since the stone is at the water surface, the final height is zero. Therefore, the final potential energy is zero. Substituting this into the equation, we get:
198J = 0 + Final kinetic energy
Final kinetic energy = 198J
Therefore, the kinetic energy of the stone just as it hits the water is 198 Joules.
D. To find the speed with which the stone hits the water, we can use the formula for kinetic energy and solve for the velocity:
Kinetic energy = 1/2 * mass * velocity^2
Substituting the given values of kinetic energy (198J) and mass (2.0kg) into the equation, we get:
198J = 1/2 * 2.0kg * velocity^2
Simplifying the equation, we have:
99J = 1.0kg * velocity^2
Rearranging the equation, we get:
velocity^2 = 99J / 1.0kg
velocity^2 = 99m^2/s^2
velocity = √99m/s
velocity ≈ 9.95m/s
Therefore, the speed with which the stone hits the water is approximately 9.95m/s.