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Physics 6
| Question | Answer |
|---|---|
| Work | the work (w) done by a force (F) that is constant in both magnitude and direction is given by the following equation: W = Fdcos0 |
| d | The displacement of the object |
| theta in work equation | The angle between the force vector and the displacement vector |
| SI unit of work | The SI unit of work is newton-meter. The newton-meter has been named the joule (J) and 1J = 1 Nm. |
| Tip about displacement and work relationship | If there's no displacement, there's no work! W = F/d |
| Tip about parallel and perpendicular | Force that is parallel does work, perpendicular doesn't. Remember that the horizontal component is the one doing work! |
| Work Done by a Variable Force | If the force acting on an object varies in magnitude and/or direction during the object's displacement, graphical analysis can be used to determine the work done. Fcos(0) is plotted on y axis and distance on x. The work is the area under the curve |
| Kinetic Energy | motion that exists only if the object is in motion. KE = 1/2mv^2. |
| Tip about relationship between kinetic energy and speed | If the speed is double, the kinetic energy is going to quadruple. Speed is more significant than the mass. |
| Gravitation Potential Energy | the energy an object possesses due to its position within a gravitational field. |
| PE = | mgh |
| Change in PE = | PEf - PEi, this will also always equal to mgh, regardless of the position. |
| Work Done By the Force of Gravity & Conservative Forces | Wmg = mg * dcos(0) or Wmg = -change in PE |
| Conservative Forces | forces where the work done does not depend upon the path taken; it only depends on the initial position and the final position. Wc = -change in PE, cause mg is a conservative force. |
| Work-Energy Theorem | the net work done on an object is equal to its change in kinetic energy. Wnet = change in kinetic energy |
| When a force makes a 90 degree angle, it will... | Cancel out and become 0 |
| Rollercoaster Demo | For a RC to reach top of loop, it needs enough mechanical energy at start. The cart has a lot of PE at the start, loses it as it goes down, and has the most KE at the bottom of the loop, which lets it maintain contact with the track throughout the loop. |
| Power | The rate at which work is done. The unit of power is the watt where 1 watt = joule/sec. |
| P = | w/t, also energy/time. If the force is constant, you can also do: F * average velocity. |
| Do Homework Problem 1 | -- |
| Do Homework Problem 2 | -- |
| Do Homework Problem 3 | -- |
| Do Homework Problem 4 | -- |
| Do Homework Problem 5 | -- |
| Do Homework Problem 6 | -- |
| Do Homework Problem 7 | -- |
| Do Homework Problem 8 | -- |
| Do Homework Problem 9 | -- |
| Do Homework Problem 10 | -- |
Created by:
smurtab