ASVAB Mechanical Comprehension Practice Test 909136 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.66
Score	0%	53%

Review

Lisa lifts a 25 pound box from the floor onto a loading dock 4 ft. off the ground. Sam slides the same box along a ramp to move it up another 4 ft. onto a flatbed truck. Who has done more work?

50% Answer Correctly

	Neither have done any work
	They have done an equal amount of work
	Lisa
	Sam

Solution

Work is force multiplied by distance. Because both Connie and Sam moved the same weight the same distance they have done an equal amount of work. Sam employed the mechnacial advantage of an inclined plane so he exerted less effort to do the work but the amount of work done was still the same.

2 If the green box weighs 30 lbs. and 65 lbs. of force is applied 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?

55% Answer Correctly

	58.5 ft.
	6.5 ft.
	19.5 ft.
	3 ft.

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:

R_ad_a = R_bd_b

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 d_a, our missing value, and plugging in our variables yields:

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{65 lbs. \times 9 ft.}{30 lbs.} \) = \( \frac{585 ft⋅lb}{30 lbs.} \) = 19.5 ft.

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 2.0 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

	333.3
	-19.8
	375
	83.3

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: F_ed_e = F_rd_r. In this problem we're looking for resistance force:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{250 \times 2.0}{1.5} \)
\( F_r = \frac{500.0}{1.5} \)
\( F_r = 333.3 \)

What type of load creates different stresses at different locations on a structure?

60% Answer Correctly

	dynamic load
	impact load
	non-uniformly distributed load
	static uniformly distributed load

Solution

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.