ASVAB Mechanical Comprehension Practice Test 466791 Results

Your Results Global Average
Questions 5 5
Correct 0 2.76
Score 0% 55%

Review

1 If A = 8 ft., B = 1 ft., C = 3 ft., the green box weighs 40 lbs. and the blue box weighs 55 lbs., what does the orange box have to weigh for this lever to balance?
43% Answer Correctly
29.44 lbs.
88.33 lbs.
40 lbs.
0 lbs.
Solution
In order for this lever to balance, the torque acting on each side of the fulrum must be equal. So, the torque produced by A must equal the torque produced by B and C. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

fAdA = fBdB + fCdC

For this problem, this equation becomes:

40 lbs. x 8 ft. = 55 lbs. x 1 ft. + fC x 3 ft.

320 ft. lbs. = 55 ft. lbs. + fC x 3 ft.

fC = \( \frac{320 ft. lbs. - 55 ft. lbs.}{3 ft.} \) = \( \frac{265 ft. lbs.}{3 ft.} \) = 88.33 lbs.

2 The green box weighs 10 lbs. and a 10 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
7 ft.
0 ft.
1 ft.
3.5 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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{10 lbs. \times 7 ft.}{10 lbs.} \) = \( \frac{70 ft⋅lb}{10 lbs.} \) = 7 ft.

3 If the handles of a wheelbarrow are 2.5 ft. from the wheel axle and the load is concentrated at a point 1.5 ft. from the axle, how many pounds of load will a 250 lbs. force lift?
47% Answer Correctly
375
416.7
None of these is correct
150
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. In this problem we're looking for resistance force:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{250 \times 2.5}{1.5} \)
\( F_r = \frac{625.0}{1.5} \)
\( F_r = 416.7 \)
4

Which of these is the formula for kinetic energy?

68% Answer Correctly

\(KE = {1 \over 2}mv^2\)

\(KE = mgh\)

\(KE = {1 \over 2}mh^2\)

\(KE = {m \over v^2 }\)
Solution

Kinetic energy is the energy of movement and is a function of the mass of an object and its speed: \(KE = {1 \over 2}mv^2\) where m is mass in kilograms, v is speed in meters per second, and KE is in joules. The most impactful quantity to kinetic energy is velocity as an increase in mass increases KE linearly while an increase in speed increases KE exponentially.

5 If the green box weighs 25 lbs. and is 9 ft. from the fulcrum, how far from the fulcrum would a 40 lbs. weight need to be placed to balance the lever?
61% Answer Correctly
16.88 ft.
1.88 ft.
5.63 ft.
225 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 db, our missing value, and plugging in our variables yields:

db = \( \frac{R_ad_a}{R_b} \) = \( \frac{25 lbs. \times 9 ft.}{40 lbs.} \) = \( \frac{225 ft⋅lb}{40 lbs.} \) = 5.63 ft.