Question

phys 100 homework set #5
1.) as water descends from the top of a tall hydroelectric dam, its gravitational potential energy is converted to electric energy. how much gravitational potential energy is released when 1000 kg of water descends 200 m to the generators?
2.) what is the speed of the water once it reaches the generators, assume it has an initial vertical velocity of zero as it begins to fall?

Explanation:

Step1: Calculate gravitational potential energy

The formula for gravitational potential energy is $PE = mgh$, where $m=1000\ \text{kg}$, $g=9.8\ \text{m/s}^2$, $h=200\ \text{m}$.
$PE = 1000 \times 9.8 \times 200$

Step2: Compute the result for PE

Calculate the product.
$PE = 1,960,000\ \text{J}$

Step3: Relate PE to kinetic energy

Assume all PE converts to KE: $KE = PE = \frac{1}{2}mv^2$, solve for $v$.
$v = \sqrt{\frac{2PE}{m}}$

Step4: Calculate final speed

Substitute values into the velocity formula.
$v = \sqrt{\frac{2 \times 1960000}{1000}} = \sqrt{3920} \approx 62.6\ \text{m/s}$

Answer:

1. The gravitational potential energy released is $1.96 \times 10^6\ \text{J}$
2. The speed of the water at the generators is approximately $62.6\ \text{m/s}$