ASVAB Mechanical Comprehension Practice Test 813801 Results

Your Results Global Average
Questions 5 5
Correct 0 2.78
Score 0% 56%

Review

1 What is the efficiency of a machine has work input of 235 ft⋅lb and work output of 199 ft⋅lb?
67% Answer Correctly
15%
85%
1%
170%
Solution
Due to friction, a machine will never be able to utilize 100% of its work input. A certain percentage of that input will be lost in overcoming friction within the machine. Effeciency is a measure of how much of a machine's work input can be turned into useful work output and is calculated by dividing work output by work input and multiplying the result by 100:
\( Efficiency = \frac{Work_{out}}{Work_{in}} \times 100 \) \( = \frac{199 ft⋅lb}{235 ft⋅lb} \times 100 \) \( = 85% \) %
2

A screw is most like which of the following other simple machines?

50% Answer Correctly

first-class lever

block and tackle

inclined plane

wheel and axle
Solution

A screw is an inclined plane wrapped in ridges (threads) around a cylinder. The distance between these ridges defines the pitch of the screw and this distance is how far the screw advances when it is turned once. The mechanical advantage of a screw is its circumference divided by the pitch.

3 If the handles of a wheelbarrow are 2.0 ft. from the wheel axle and the load is concentrated at a point 1.0 ft. from the axle, how many pounds of load will a 270 lbs. force lift?
47% Answer Correctly
0
270
1003.2
540
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:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{270 \times 2.0}{1.0} \)
\( F_r = \frac{540.0}{1.0} \)
\( F_r = 540 \)
4

Coplanar forces:

62% Answer Correctly

pass through a common point

act in a common plane

act along the same line of action

have opposite dimensions
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.

5 How much work can a 8 hp engine do in 10 seconds?
52% Answer Correctly
44000 ft⋅lb
32 ft⋅lb
1 ft⋅lb
4 ft⋅lb
Solution
Horsepower (hp) is a common measure of power output for complex machines. By definition, a 1 hp machine does 550 ft⋅lb of work in 1 second: 1 hp = 550 ft⋅lb/s. Substituting the variables for this problem gives us:
\( W = 8 hp \times 550 \frac{ft⋅lb}{s} \times 10s = 44000 ft⋅lb \)