ASVAB Mechanical Comprehension Practice Test 568553 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.35
Score	0%	67%

Review

1 If A = 8 ft. and the green box weighs 50 lbs. what is the torque acting on the A side of this lever?

76% Answer Correctly

	200 ft⋅lb
	400 ft⋅lb
	100 ft⋅lb
	6 ft⋅lb

Solution

For a lever, torque is weight x distance from the fulcrum which, in this case, is: 50 ft. x 8 lbs. = 400 ft⋅lb

Power is the rate at which:

63% Answer Correctly

	friction is overcome
	input force is transferred to output force
	potential energy is converted into kinetic energy
	work is done

Solution

Power is the rate at which work is done, P = w/t, or work per unit time. The watt (W) is the unit for power and is equal to 1 joule (or newton-meter) per second. Horsepower (hp) is another familiar unit of power used primarily for rating internal combustion engines. 1 hp equals 746 watts.

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

53% Answer Correctly

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

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.

A ramp is an example of which kind of simple machine?

85% Answer Correctly

	inclined plane
	wedge
	none of these
	first-class lever

Solution

An inclined plane is a simple machine that reduces the force needed to raise an object to a certain height. Work equals force x distance and, by increasing the distance that the object travels, an inclined plane reduces the force necessary to raise it to a particular height. In this case, the mechanical advantage is to make the task easier. An example of an inclined plane is a ramp.

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

	810 lbs
	0.83 lbs
	45 lbs
	90 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