| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.02 |
| Score | 0% | 60% |
On Earth, acceleration due to gravity (g) is approximately __________.
6.67 x 10-11 m/s2 |
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9.8 m/s2 |
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1 m/s |
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1 m/s2 |
Newton's Law of Univeral Gravitation defines the general formula for the attraction of gravity between two objects: \(\vec{F_{g}} = { Gm_{1}m_{2} \over r^2}\) . In the specific case of an object falling toward Earth, the acceleration due to gravity (g) is approximately 9.8 m/s2.
Assuming force applied remains constant, which of the following will result in more work being done?
moving the object with more speed |
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moving the object farther |
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moving the object with more acceleration |
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increasing the coefficient of friction |
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.
| 22.86 lbs. | |
| 105 lbs. | |
| 11 lbs. | |
| 7.62 lbs. |
fAdA = fBdB + fCdC
For this problem, this equation becomes:
35 lbs. x 8 ft. = 40 lbs. x 3 ft. + fC x 7 ft.
280 ft. lbs. = 120 ft. lbs. + fC x 7 ft.
fC = \( \frac{280 ft. lbs. - 120 ft. lbs.}{7 ft.} \) = \( \frac{160 ft. lbs.}{7 ft.} \) = 22.86 lbs.
What is the first step to solving a problem where multiple forces are acting on an object?
calculate kinetic energy |
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calculate the total force |
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calculate the net force |
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calculate potential energy |
In mechanics, multiple forces are often acting on a particular object and, taken together, produce the net force acting on that object. Like force, net force is a vector quantity in that it has magnitude and direction.
| 26.22 lbs. | |
| 3 lbs. | |
| 8 lbs. | |
| 11 lbs. |
The mechanical advantage of a wheel and axle is the input radius divided by the output radius:
MA = \( \frac{r_i}{r_o} \)
In this case, the input radius (where the effort force is being applied) is 8 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{8}{3} \) = 2.67
MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{70 lbs.}{2.67} \) = 26.22 lbs.