ASVAB Mechanical Comprehension Practice Test 968557 Results

Your Results Global Average
Questions 5 5
Correct 0 2.93
Score 0% 59%

Review

1 If the green box weighs 20 lbs. and is 9 ft. from the fulcrum, how much weight would need to be placed at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 6 ft.?
63% Answer Correctly
10 lbs.
30 lbs.
60 lbs.
3 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 Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{20 lbs. \times 9 ft.}{6 ft.} \) = \( \frac{180 ft⋅lb}{6 ft.} \) = 30 lbs.

2

Friction between two or more solid objects that are not moving relative to each other is called:

73% Answer Correctly

dynamic friction

gravitational friction

static friction

kinetic friction
Solution

Static friction is friction between two or more solid objects that are not moving relative to each other. An example is the friction that prevents a box on a sloped surface from sliding farther down the surface.

3

Which of these will have the most impact on the kinetic energy of an object?

54% Answer Correctly

its weight

its mass

its speed

its direction
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.

4 If the handles of a wheelbarrow are 2.5 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 140 lbs. force lift?
47% Answer Correctly
17.1
None of these is correct
119.9
350
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:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{140 \times 2.5}{1.0} \)
\( F_r = \frac{350.0}{1.0} \)
\( F_r = 350 \)
5

For any given surface, the coefficient of static friction is ___________ the coefficient of kinetic friction.

54% Answer Correctly

opposite

higher than

lower than

equal to
Solution

For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).