As more and more renewable energy is being used to provide p

As more and more renewable energy is being used to provide power to the grid, the problem of how to store the energy for later use has become important. Solar plants for example only generate power during the day, and wind turbines only generate power when the wind is blowing. To provide a steady source of energy, these plants must be paired with energy storage. Batteries are one option, but they tend to be relatively expensive. Recently a technology called Advanced Rail Energy Storage (ARES) has been developed and tested in Tehachapi, California (http://www.aresnorthamerica.com). The concept is simple: An electric freight train pulls cars loaded with massive concrete blocks up the side of a mountain using the excess power produced by a nearby renewable power plant. At the top the blocks can be unloaded. When more energy is needed by the grid than the power plant can provide, the blocks are loaded on the train at the top and the train is allowed to roll down, with the electric engine acting in reverse as a generator much the way the regenerative braking on a hybrid car works to deliver power to the battery. According to the ARES press kit the baseline system has been designed to provide 333 MW of power for up to 8 hours using tracks at a 7.5% grade that are 8 miles in length. The system has thousands of concrete blocks each with a weight of 240 tons. At the top and bottom of the track are switcher yards that allow multiple trains to park simultaneously.

3. When the engine acts as a regenerative brake it prevents the train from accelerating as it rolls down the hill - converting potential energy into electrical energy rather than kinetic energy. What braking force does this provide for a train weighing 1200 tons?

Solution

1ton = 1000kg

Therefore, mass m = 1200 ton = 1200 * 1000kg = 1.2 x 10^6 kg

Let d be the vertical distance travelled down the hill.

Then the work done by the braking force is given by:

W = change in potential energy = mgd = (1.2 x 10^6)(9.81)d J

Now,

Braking force F = W/d = (1.2 x 10^6)(9.81)d/d = 11,772,000 N