ASVAB Mechanical Comprehension Practice Test 300367 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.08
Score	0%	62%

Review

Which of these is the formula for kinetic energy?

67% Answer Correctly

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

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 box weighs 55 lbs. and is 5 ft. from the fulcrum, how far from the fulcrum would a 15 lbs. force need to be applied to balance the lever?

58% Answer Correctly

	36.67 ft.
	11 ft.
	73.33 ft.
	18.33 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_b, our missing value, and plugging in our variables yields:

d_b = \( \frac{R_ad_a}{R_b} \) = \( \frac{55 lbs. \times 5 ft.}{15 lbs.} \) = \( \frac{275 ft⋅lb}{15 lbs.} \) = 18.33 ft.

3 What is the efficiency of a machine has work input of 235 ft⋅lb and work output of 94 ft⋅lb?

67% Answer Correctly

	40%
	80%
	2%
	0%

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:
\( Efficiency = \frac{Work_{out}}{Work_{in}} \times 100 \) \( = \frac{94 ft⋅lb}{235 ft⋅lb} \times 100 \) \( = 40% \) %

Which of the following will increase the mechanical advantage of a second-class lever?

55% Answer Correctly

	decrease the length of the lever
	move the object being lifted closer to the fulcrum
	move the object being lifted farther away from the fulcrum
	move the fulcrum between the force and the object being lifted

Solution

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.

Normal force is generally equal to the __________ of an object.

61% Answer Correctly

	weight
	mass
	coefficient of friction
	density

Solution

Normal force arises on a flat horizontal surface in response to an object's weight pressing it down. Consequently, normal force is generally equal to the object's weight.