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# Kinetic Energy

Energy is the currency of transformation. All systems remain the same unless energy is exchanged between one part and another. Learn here about the energy associated with movement.

Hydrovolts has invented a water powered cell phone charger - the HydroBee. Put the charger in a flow of water, like a stream or faucet, and a little turbine spins, generates electricity, and charges up your phone. Hence if you are out camping you can now charge a phone without carrying along a bunch of batteries. If the HydroBee has a circular opening aperture of radius \(5~\mbox{cm}\) and is placed in a stream with flow rate \(1~\mbox{m/s}\), how long **in hours** will it take to completely charge an empty cell phone battery rated at \(5~\mbox{W-h}\) (Watt-hours)?

**Details and assumptions**

- The density of water is \(\rho=1~\mbox{g/cm}^3\).
- The energy conversion efficiency of the kinetic energy of the water to electric energy is 50%.
- The water flows through the entire opening aperture.

The caber toss is a traditional Scottish sport that involves hurling a caber, which is essentially a large piece of a tree: a caber is a \(6~\mbox{m}\) long, \(80~\mbox{kg}\) tree trunk. Large strong people hurl these and the goal is to get the caber to land as far away as possible and rotate in the air, so what was the highest part of the trunk initially is actually the part that hits the gound first. See this clip to understand how the caber rotates.

It's HARD to do this. To see how hard, consider a perfectly vertical caber with one end on the ground. You then launch the caber vertically with some speed \(v_0\) and give it a rotation. What is the minimum kinetic energy **in Joules** you need to give the caber so that when it lands the caber is perfectly vertical again, but the OTHER end of the caber hits the ground?

**Details and assumptions**

- The acceleration of gravity is \(-9.8~\mbox{m/s}^2\).
- The caber can be modeled as a uniform rod.

**in meters per second** of the charge located at C.

A bead slides down on a frictionless wire, starting from rest at point A. The section from A to B is straight while the section from B to C is wavy (with wavelength \(\lambda\ll L\)). The total length of the stretched wire is \(3L\). If \(L=50~\text{cm}\), determine the time **in seconds** it takes for the bead to reach point C. Assume that the bead slides **smoothly** at all times.

**Details and assumptions**

\[g=9.8~\text{m}/\text{s}^2\]

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