| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.08 |
| Score | 0% | 62% |
Which of the following is the formula for gravitational potential energy?
\(PE = { 1 \over 2} mv^2\) |
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\(PE = mgh\) |
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\(PE = mg^2h\) |
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\(PE = { 1 \over 2} mg^2\) |
Gravitational potential energy is energy by virtue of gravity. The higher an object is raised above a surface the greater the distance it must fall to reach that surface and the more velocity it will build as it falls. For gravitational potential energy, PE = mgh where m is mass (kilograms), h is height (meters), and g is acceleration due to gravity which is a constant (9.8 m/s2).
| 15.47 lbs. | |
| 185.63 lbs. | |
| 61.88 lbs. | |
| 0 lbs. |
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 Rb, our missing value, and plugging in our variables yields:
Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{55 lbs. \times 9 ft.}{8 ft.} \) = \( \frac{495 ft⋅lb}{8 ft.} \) = 61.88 lbs.
Normal force is generally equal to the __________ of an object.
coefficient of friction |
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density |
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mass |
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weight |
Normal force arises on a flat horizontal surface in response to an object's weight pressing it down. Consequently, normal force is generally equal to the object's weight.
What type of load is sudden and for a relatively short duration?
dynamic load |
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concentrated load |
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impact load |
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non-uniformly distributed load |
A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.
| 100 lbs. | |
| 33.3 lbs. | |
| 30 lbs. | |
| 35.3 lbs. |
This problem describes an inclined plane and, for an inclined plane, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance:
Fede = Frdr
Plugging in the variables from this problem yields:
Fe x 15 ft. = 500 lbs. x 1 ft.
Fe = \( \frac{500 ft⋅lb}{15 ft.} \) = 33.3 lbs.