ASVAB Mechanical Comprehension Practice Test 801486 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.06
Score	0%	61%

Review

The work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. This defines which of the following?

60% Answer Correctly

	mechanical advantage
	Pascal's law
	conservation of mechanical energy
	work-energy theorem

Solution

The work-energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. Simply put, work imparts kinetic energy to the matter upon which the work is being done.

2 The green box weighs 5 lbs. and a 5 lbs. weight is placed 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?

57% Answer Correctly

	9 ft.
	36 ft.
	45 ft.
	27 ft.

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

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{5 lbs. \times 9 ft.}{5 lbs.} \) = \( \frac{45 ft⋅lb}{5 lbs.} \) = 9 ft.

3 90 lbs. of effort is used by a machine to lift a 720 lbs. box. What is the mechanical advantage of the machine?

84% Answer Correctly

	8.8
	8
	16
	10

Solution

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{720 lbs.}{90 lbs.} \) = 8

4 If the handles of a wheelbarrow are 0.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 180 lbs. load concentrated at a point 1.0 ft. from the axle?

52% Answer Correctly

	592.8
	None of these is correct
	41
	360

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 effort force:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{180 \times 1.0}{0.5} \)
\( F_e = \frac{180.0}{0.5} \)
\( F_e = 360 \)

Assuming force applied remains constant, which of the following will result in more work being done?

53% Answer Correctly

	moving the object with more acceleration
	moving the object with more speed
	moving the object farther
	increasing the coefficient of friction

Solution

Work is accomplished when force is applied to an object: W = Fd where F is force in newtons (N) and d is distance in meters (m). Thus, the more force that must be applied to move an object, the more work is done and the farther an object is moved by exerting force, the more work is done.