ASVAB Mechanical Comprehension Practice Test 702735 Results

Your Results Global Average
Questions 5 5
Correct 0 3.16
Score 0% 63%

Review

1

Assuming force applied remains constant, which of the following will result in more work being done?

53% Answer Correctly

moving the object with more acceleration

moving the object with more speed

moving the object farther

increasing the coefficient of friction
Solution

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.

2

Collinear forces:

73% Answer Correctly

act along the same line of action

are unrelated to each other

act in a common plane

pass through a common point
Solution

Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.

3

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

90 lbs

45 lbs

810 lbs

0.83 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

4 If the green box weighs 70 lbs. and is 3 ft. from the fulcrum, how much force would need to be applied at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 6 ft.?
62% Answer Correctly
105 lbs.
35 lbs.
11 lbs.
8.75 lbs.
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 Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{70 lbs. \times 3 ft.}{6 ft.} \) = \( \frac{210 ft⋅lb}{6 ft.} \) = 35 lbs.

5 If you lift a 34 lbs. rock 16 ft. from the ground, how much work have you done?
71% Answer Correctly
1088 ft⋅lb
2 ft⋅lb
0 ft⋅lb
544 ft⋅lb
Solution
Work is force times distance. In this case, the force is the weight of the rock so:
\( W = F \times d \)
\( W = 34 \times 16 \)
\( W = 544 \)