ASVAB Mechanical Comprehension Practice Test 694330 Results

Your Results Global Average
Questions 5 5
Correct 0 3.34
Score 0% 67%

Review

1 If you lift a 13 lbs. rock 21 ft. from the ground, how much work have you done?
70% Answer Correctly
0 ft⋅lb
1 ft⋅lb
273 ft⋅lb
34 ft⋅lb
Solution
Work is force times distance. In this case, the force is the weight of the rock so:
\( W = F \times d \)
\( W = 13 \times 21 \)
\( W = 273 \)
2

On Earth, acceleration due to gravity (g) is approximately __________. 

81% Answer Correctly

6.67 x 10-11 m/s2

9.8 m/s2

1 m/s

1 m/s2
Solution

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. 

3 The green box weighs 20 lbs. and a 70 lbs. weight is placed 9 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
94.5 ft.
31.5 ft.
7.88 ft.
0 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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{70 lbs. \times 9 ft.}{20 lbs.} \) = \( \frac{630 ft⋅lb}{20 lbs.} \) = 31.5 ft.

4 If a 60 lbs. weight is placed 7 ft. from the fulcrum at the blue arrow and the green box is 6 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?
61% Answer Correctly
140 lbs.
210 lbs.
70 lbs.
35 lbs.
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 Ra, our missing value, and plugging in our variables yields:

Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{60 lbs. \times 7 ft.}{6 ft.} \) = \( \frac{420 ft⋅lb}{6 ft.} \) = 70 lbs.

5

If the handles of a wheelbarrow are 3 ft. from the wheel axle, what force must you exert to lift the handles if it's carrying a 270 lb. load concentrated at a point 0.5 ft. from the axle?

56% Answer Correctly

0.83 lbs

90 lbs

45 lbs

810 lbs
Solution

This problem describes a second-class lever and, for a second class lever, 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 3 ft. = 270 lbs x 0.5 ft
Fe = 135 ft-lb. / 3 ft 
Fe = 45 lbs