ASVAB Mechanical Comprehension Practice Test 706398 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.05
Score	0%	61%

Review

Which of these is the formula for kinetic energy?

68% Answer Correctly

	\(KE = mgh\)
	\(KE = {m \over v^2 }\)
	\(KE = {1 \over 2}mv^2\)
	\(KE = {1 \over 2}mh^2\)

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.

2 If the green arrow in this diagram represents 600 ft⋅lb of work, how far will the box move if it weighs 150 pounds?

73% Answer Correctly

	4 ft.
	37 ft.
	2 ft.
	16 ft.

Solution

The Law of Work states that the work put into a machine is equal to the work received from the machine under ideal conditions. In equation form, that's:

W_in = W_out
F_effort x d_effort = F_resistance x d_resistance

In this problem, the effort work is 600 ft⋅lb and the resistance force is 150 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
600 ft⋅lb = 150 lbs. x d_resistance
d_resistance = \( \frac{600ft⋅lb}{150 lbs.} \) = 4 ft.

Depending on where you apply effort and resistance, the wheel and axle can multiply:

45% Answer Correctly

	force or distance
	force or speed
	power or distance
	speed or power

Solution

If you apply the resistance to the axle and the effort to the wheel, the wheel and axle will multiply force and if you apply the resistance to the wheel and the effort to the axle, it will multiply speed.

4 If the green box weighs 10 lbs. and is 7 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

	0 lbs.
	23.33 lbs.
	5.83 lbs.
	11.67 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:

R_ad_a = R_bd_b

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

R_b = \( \frac{R_ad_a}{d_b} \) = \( \frac{10 lbs. \times 7 ft.}{6 ft.} \) = \( \frac{70 ft⋅lb}{6 ft.} \) = 11.67 lbs.

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
	0.83 lbs
	45 lbs
	810 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: F_ed_e = F_rd_r. Plugging in the variables from this problem yields:

F_e x 3 ft. = 270 lbs x 0.5 ft
F_e = 135 ft-lb. / 3 ft
F_e= 45 lbs