# Energy from fall object

1. Apr 2, 2014

### Mryeh

Hi guys,

I have a question that has been intriguing me for some time.

If i drop an object 100m at the speed of gravity (9.8), which has a mass of 1000kg - given the initial velocity of zero, i get the following results:

velocity = 44.3 m/s
Kinetic energy = 980000 Joules

If my calculations are correct, would the object generate 435kw of energy given that it takes about 2.4 seconds to fall?

Notwithstanding any loss of energy, can this motion energy be transformed into electrical energy via a turbine generator?

Thanks.

2. Apr 2, 2014

### ZapperZ

Staff Emeritus
I'm not exactly sure how you calculated the power, considering that the amount of KE during the fall isn't a constant (the velocity isn't a constant). Why can't you use calculate the gravitational potential energy?

Secondly, isn't this the concept being used at hydroelectric dam? A relatively constant amount of water from some height, possessing gravitational potential energy, is used to turn turbines to generate electricity. Water fall is a "falling object".

Zz.

Last edited: Apr 2, 2014
3. Apr 2, 2014

### Staff: Mentor

The power is determined by the amount of time it takes to "catch" the object and slow it from 44.3m/s to zero speed, not the time it takes to fall from its starting point. If the "stopping time" is 1s, you get 235kw of power, that lasts for 1s. If it's 0.5s, you get 470kw that lasts 0.5s.

If you drop one object every 10s, and it takes 0.5s to catch each one, you get a series of 470kw spikes, each 0.5s long, that average to 23.5kw over the entire 10s. If you want to get a steady 23.5kw, you need a way to store the energy and release it at the correct rate.

[added: as ZapperZ notes, it's easier to get a steady rate of power production if you let the mass fall in a continuous stream (e.g. water in a hydroelectric dam) rather in 1000kg chunks.]

4. Apr 2, 2014

### Mryeh

Thanks for the replies. Catching the object and slowing it down would apply if if the object would hit the turbine as in the hydro-electric system. But what if the object is connect to a rope which in turn spins the turbine as the object drops - hence the relevance of the time. Jtbell, would this effect the figures you provided?

5. Apr 2, 2014

### nasu

If it's attached to the rope and wheel the calculation of speed, acceleration and times is not valid anymore. There will be a force acting on the weight from the rope.
The overall energy is still the same if you neglect dissipation.

Hydroelectric plants are still better than weights on ropes. You don't need to lift the water back on top. The sun does this for free.

6. Apr 2, 2014

### Mryeh

Thanks Nasu. To answer zapper z, i arrived at the power via the following steps:

- velocity: v = √2gh = √2x9.8x100 = 44.27
- Kinetic energy: KE = 1/2mv² = 1/2 x 1000 x 44.27² = 980000J
- Power (w): P = E/t = w = J/s w = 980000 / 2.25 = 435555.5w (435.5KW)

Are my calculations correct?

Nasu, withstanding the rope and the wheel, is the object generating 435.5KW?

7. Apr 2, 2014

### ZapperZ

Staff Emeritus
No it is not. The "v" that you used is NOT a constant over the time period. Besides, you are saying that over that whole 2.25 seconds, it can do this amount of work per second. That is wrong, because the KE is that you calculated is only correct at the BOTTOM.

Be careful of plugging and chugging away without understanding what you are doing.

This scheme that you are discussing doesn't work. The "work" done is only "instantaneous", not constant, and not as an average.

Zz.

8. Apr 2, 2014

### Staff: Mentor

Then you still get a total 235kJ of energy from the "falling" object (the difference in its potential energy at the top versus bottom), neglecting losses due to friction. The power depends on the rate (speed) at which you lower the object. If you lower the object at constant speed and take 1s to do it, you get a constant 235kW of power for 1s. If you take 10s to lower the object (again at constant speed) you get a constant 23.5kW of power for 10s. If the speed varies, then the instantaneous power also varies.

9. Apr 2, 2014

### sophiecentaur

If you are initially assuming that the system is 100% efficient (which you seem to be doing), then you do not need to calculate KE. The gravitational Potential Energy at the top per second (using the mass arriving per second and mgh) will be equal to the Power delivered at the bottom.
If you are 'lowering the object down' by braking, then, of course, you are losing energy and the calculations become harder.