| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.76 |
| Score | 0% | 55% |
Power is the rate at which:
potential energy is converted into kinetic energy |
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input force is transferred to output force |
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friction is overcome |
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work is done |
Power is the rate at which work is done, P = w/t, or work per unit time. The watt (W) is the unit for power and is equal to 1 joule (or newton-meter) per second. Horsepower (hp) is another familiar unit of power used primarily for rating internal combustion engines. 1 hp equals 746 watts.
What's the last gear in a gear train called?
driver gear |
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driven gear |
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output gear |
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idler gear |
A gear train is two or more gears linked together. Gear trains are designed to increase or reduce the speed or torque outpout of a rotating system or change the direction of its output. The first gear in the chain is called the driver and the last gear in the chain the driven gear with the gears between them called idler gears.
A a seesaw / teeter-totter is an example of which of the following?
first-class lever |
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inclined plane |
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second-class lever |
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third-class lever |
A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter.
| 1.3 ft. | |
| 0 ft. | |
| 2.6 ft. | |
| 3.9 ft. |
To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.Solving for db, our missing value, and plugging in our variables yields:
db = \( \frac{R_ad_a}{R_b} \) = \( \frac{65 lbs. \times 1 ft.}{50 lbs.} \) = \( \frac{65 ft⋅lb}{50 lbs.} \) = 1.3 ft.
Assuming force applied remains constant, which of the following will result in more work being done?
moving the object farther |
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increasing the coefficient of friction |
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moving the object with more speed |
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moving the object with more acceleration |
Work is accomplished when force is applied to an object: W = Fd where F is force in newtons (N) and d is distance in meters (m). Thus, the more force that must be applied to move an object, the more work is done and the farther an object is moved by exerting force, the more work is done.