ASVAB Mechanical Comprehension Practice Test 95935 Results

Your Results Global Average
Questions 5 5
Correct 0 3.13
Score 0% 63%

Review

1 A 210 lb. barrel is rolled up a 10 ft. ramp to a platform that's 3 ft. tall. What effort is required to move the barrel?
53% Answer Correctly
69.3 lbs.
63 lbs.
64.5 lbs.
66 lbs.
Solution

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 10 ft. = 210 lbs. x 3 ft.
Fe = \( \frac{630 ft⋅lb}{10 ft.} \) = 63 lbs.

2

One Horsepower (hp) is equal to how many watts?

76% Answer Correctly

1

746

9.8

1492
Solution

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.

3 If the green box weighs 50 lbs. and 50 lbs. of force is applied 7 ft. from the fulcrum at the blue arrow, how far from the fulcrum would the green box need to be placed to balance the lever?
55% Answer Correctly
28 ft.
2.33 ft.
7 ft.
1.75 ft.
Solution

To balance this lever the torques at the green box and the blue arrow must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.

Solving for da, our missing value, and plugging in our variables yields:

da = \( \frac{R_bd_b}{R_a} \) = \( \frac{50 lbs. \times 7 ft.}{50 lbs.} \) = \( \frac{350 ft⋅lb}{50 lbs.} \) = 7 ft.

4 If you lift a 6 lbs. rock 5 ft. from the ground, how much work have you done?
70% Answer Correctly
30 ft⋅lb
None of these is correct
15 ft⋅lb
60 ft⋅lb
Solution
Work is force times distance. In this case, the force is the weight of the rock so:
\( W = F \times d \)
\( W = 6 \times 5 \)
\( W = 30 \)
5 The green box weighs 10 lbs. and a 50 lbs. weight is placed 7 ft. from the fulcrum at the blue arrow. How far from the fulcrum would the green box need to be placed to balance the lever?
57% Answer Correctly
105 ft.
17.5 ft.
8.75 ft.
35 ft.
Solution

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 da, our missing value, and plugging in our variables yields:

da = \( \frac{R_bd_b}{R_a} \) = \( \frac{50 lbs. \times 7 ft.}{10 lbs.} \) = \( \frac{350 ft⋅lb}{10 lbs.} \) = 35 ft.