ASVAB Mechanical Comprehension Practice Test 546578 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.28
Score	0%	66%

Review

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

55% Answer Correctly

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

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.

2 60 lbs. of effort is used by a machine to lift a 180 lbs. box. What is the mechanical advantage of the machine?

84% Answer Correctly

	2.7
	3.3
	4.5
	3

Solution

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{180 lbs.}{60 lbs.} \) = 3

3 If A = 12 ft., B = 3 ft., C = 7 ft., the green box weighs 25 lbs. and the blue box weighs 40 lbs., what does the orange box have to weigh for this lever to balance?

43% Answer Correctly

	8.57 lbs.
	77.14 lbs.
	25.71 lbs.
	2 lbs.

Solution

In order for this lever to balance, the torque acting on each side of the fulrum must be equal. So, the torque produced by A must equal the torque produced by B and C. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

f_Ad_A = f_Bd_B + f_Cd_C

For this problem, this equation becomes:

25 lbs. x 12 ft. = 40 lbs. x 3 ft. + f_C x 7 ft.

300 ft. lbs. = 120 ft. lbs. + f_C x 7 ft.

f_C = \( \frac{300 ft. lbs. - 120 ft. lbs.}{7 ft.} \) = \( \frac{180 ft. lbs.}{7 ft.} \) = 25.71 lbs.

Which of the following is the formula for gravitational potential energy?

61% Answer Correctly

	\(PE = { 1 \over 2} mv^2\)
	\(PE = { 1 \over 2} mg^2\)
	\(PE = mg^2h\)
	\(PE = mgh\)

Solution

Gravitational potential energy is energy by virtue of gravity. The higher an object is raised above a surface the greater the distance it must fall to reach that surface and the more velocity it will build as it falls. For gravitational potential energy, PE = mgh where m is mass (kilograms), h is height (meters), and g is acceleration due to gravity which is a constant (9.8 m/s²).

Friction resists movement in a direction __________ to the movement.

81% Answer Correctly

	normal
	perpendicular
	opposite
	parallel

Solution

Friction resists movement. Kinetic (also called sliding or dynamic) friction resists movement in a direction opposite to the movement. Because it opposes movement, kinetic friction will eventually bring an object to a stop. An example is a rock that's sliding across ice.