A 7 kilogram bowling ball is dropped from a fourth story window. If each story is 3 meters high, use energy conservation (not the equations of motion) to find the ball's velocity when it strikes the ground. Ignore air resistance.
Thank you
To find the velocity of the bowling ball when it strikes the ground using energy conservation, we need to consider the potential and kinetic energies involved in the system.
The potential energy (PE) of an object at a certain height is given by the product of its mass (m), the acceleration due to gravity (g), and the height (h):
PE = m * g * h
In this case, the height of the window is three stories, which is equivalent to 3 meters. So, the initial potential energy of the bowling ball when it is dropped is:
PE_initial = m * g * h_initial
Where m = 7 kg, g = 9.8 m/s^2 (acceleration due to gravity), and h_initial = 3 meters.
The equation for kinetic energy (KE) is given by:
KE = (1/2) * m * v^2
Where v represents the velocity of the ball.
Using energy conservation, we know that the total mechanical energy of the system is conserved, which means that the initial gravitational potential energy will be converted into kinetic energy when the ball strikes the ground. Thus, we can equate the initial potential energy to the final kinetic energy:
PE_initial = KE_final
Substituting the equations for potential energy and kinetic energy, we have:
m * g * h_initial = (1/2) * m * v^2
Simplifying the equation, we can cancel out the mass (m) on both sides:
g * h_initial = (1/2) * v^2
Now, we can solve for the velocity (v) by rearranging the equation:
v^2 = 2 * g * h_initial
Taking the square root of both sides:
v = √(2 * g * h_initial)
Substituting the known values, we have:
v = √(2 * 9.8 * 3) m/s
v ≈ 7.67 m/s
Therefore, the velocity of the bowling ball when it strikes the ground is approximately 7.67 m/s.