How to Calculate Potential Energy and Power Output of a Falling Mass

[pmfinnegan2]

If anyone could help me with my homework I would GREATLY appreciate it. I am totally lost. There are 2 questions. If you could tell me what/how to do it, I'd be happy to do it. I just need to know how. Thanks so much

Here is question 2:

A 5.5 kg mass is dropped at a constant velocity thru a vertical displacement of 2.0 m over a time interval of 4.0 seconds.

How much potential energy (joule) has it given up?
What is the power (watt) output?

If that power output was kept up for 24 hours how many kWhr of energy would be produced?

 

[LeonhardEuler]

"How much potential energy (joule) has it given up?"

Gravitational potential energy is equal to mgh, so the change is $$\ mg \Delta\ h$$.

"What is the power (watt) output?"

The power is the time rate of change in energy. Since the forces are constant throughout the process, this can be found by dividing the total energy change by the length of time.

"If that power output was kept up for 24 hours how many kWhr of energy would be produced?"

This is just multiplying the power output by the time, but paying attention to units.

 

[pmfinnegan2]

"How much potential energy (joule) has it given up?"

Gravitational potential energy is equal to mgh, so the change is $$\ mg \Delta\ h$$.

"What is the power (watt) output?"

The power is the time rate of change in energy. Since the forces are constant throughout the process, this can be found by dividing the total energy change by the length of time.

"If that power output was kept up for 24 hours how many kWhr of energy would be produced?"

This is just multiplying the power output by the time, but paying attention to units.

Ok, based on what you are telling me I came up with:
How much potential energy (joule) has it given up? = 107.8 J
What is the power (watt) output? = 26.95 W
If that power output was kept up for 24 hours how many kWhr of energy would be produced? = 646.8 kWh

Am I doing it correctly?

 

[LeonhardEuler]

Pretty much correct, but you didn't switch to kilowatts, so your last answer is off by a factor of 1000.

 

[pmfinnegan2]

Pretty much correct, but you didn't switch to kilowatts, so your last answer is off by a factor of 1000.

Thank you, I did forget to do that.