One kg of water is moving 2 m/s as it goes over a dam. At the base of the dam, 13m below, how fast is that water going if air friction can be neglected? (ans: 16.25m/s)
B: if the water does not go straight over the dam, but instead goes through the hydroelectric generator in the base of the dam, creating 100 J of electrical potential energy in the process, how fast will it be going after it exits the power plant at the bottom? (ans 8 m/s)
I got to figure out how they got the answers...
final Kenergy=initial PE + initial KE
solve for Kenergy, then solve for velocity.
On the second, just subtract 100J from final KE, then solve for velocity
To find the solution to both parts of the question, we can apply the principle of conservation of energy. The principle states that energy is neither created nor destroyed but rather transferred from one form to another.
In the first scenario, where air friction can be neglected, we can use the principle to equate the initial kinetic energy of the water to the sum of its potential energy and final kinetic energy:
Initial Kinetic Energy = Potential Energy + Final Kinetic Energy
The initial kinetic energy of the water is given by the formula for kinetic energy:
KE = (1/2) * m * v^2
where m is the mass of water (1 kg) and v is the velocity of the water (2 m/s).
Initially, the water only possesses kinetic energy, so the initial kinetic energy is:
KE_initial = (1/2) * 1 kg * (2 m/s)^2 = 2 J
The potential energy at the base of the dam is given by:
PE = m * g * h
where g is the acceleration due to gravity (9.8 m/s^2) and h is the height difference between the two points (13 m).
The potential energy at the base of the dam is:
PE_base = 1 kg * 9.8 m/s^2 * 13 m = 127.4 J
Now, let's assume that the final velocity of the water at the base of the dam is v_base. Therefore, the final kinetic energy can be written as:
KE_base = (1/2) * 1 kg * v_base^2
Now we can rewrite the conservation of energy equation:
KE_initial = PE_base + KE_base
Substituting the values we calculated:
2 J = 127.4 J + (1/2) * 1 kg * v_base^2
Rearranging the equation and solving for v_base:
v_base^2 = (2 J - 127.4 J) / (1/2) * 1 kg
v_base^2 = -252.8 J / -0.5 kg
v_base^2 = 505.6 m^2/s^2
v_base ≈ 22.5 m/s
Hence, the speed of the water at the base of the dam is approximately 22.5 m/s.
In the second scenario, with the hydroelectric generator in the base of the dam, we need to account for the electrical potential energy generated.
The electrical potential energy (EPE) can be calculated using the formula:
EPE = q * V
where q is the charge (in Coulombs) and V is the voltage (in Volts). Since the question states that the electrical potential energy created is 100 J, we don't have the charge or voltage values explicitly. However, we can still proceed to find the change in kinetic energy.
The change in kinetic energy is given by:
ΔKE = KE_final - KE_initial
Substituting the values:
ΔKE = (1/2) * 1 kg * v_base^2 - 2 J
However, we are interested in the final velocity after the water exits the power plant at the bottom, denoted as v_exit.
So, let's assume that the final kinetic energy is:
KE_exit = (1/2) * 1 kg * v_exit^2
We can rewrite the equation for the change in kinetic energy:
ΔKE = KE_exit - KE_initial
Substituting the values:
(1/2) * 1 kg * v_exit^2 - 2 J = 100 J
Rearranging the equation and solving for v_exit:
(1/2) * 1 kg * v_exit^2 = 102 J
v_exit^2 = 204 J / (1/2) kg
v_exit^2 = 408 m^2/s^2
v_exit ≈ 20.2 m/s
Hence, the speed of the water after it exits the power plant at the bottom is approximately 20.2 m/s.