ASVAB Mechanical Comprehension Practice Test 309812 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.21
Score	0%	64%

Review

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

	45 lbs
	810 lbs
	90 lbs
	0.83 lbs

Solution

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

Sam can do 50 ft. lb. of work in 2 minutes and 5 seconds. What would Sam have to do to increase his power output?

64% Answer Correctly

	do 100 ft. lb. of work in 4 minutes 12 seconds
	do 25 ft. lb. of work in 2 minutes 5 seconds
	do the work in 3 minutes
	do the work in 2 minutes

Solution

Power is the rate of doing work or \(\frac{W}{t}\). To increase power, increase the work being done in the same amount of time or do the same amount of work in less time.

The force exerted on an object due to gravity is called:

70% Answer Correctly

	density
	weight
	potential energy
	mass

Solution

Mass is an intrinsic property of matter and does not vary. Weight is the force exerted on the mass of an object due to gravity and a specific case of Newton's Second Law of Motion. Replace force with weight and acceleration with acceleration due to gravity on Earth (g) and the result is the formula for weight: W = mg or, substituting for g, weight equals mass multiplied by 9.8 m/s².

The mechanical advantage of connected gears is proportional to which characteristic of the gears?

73% Answer Correctly

	circumference
	speed
	number of teeth
	diameter

Solution

The mechanical advantage (amount of change in speed or torque) of connected gears is proportional to the number of teeth each gear has. Called gear ratio, it's the ratio of the number of teeth on the larger gear to the number of teeth on the smaller gear. For example, a gear with 12 teeth connected to a gear with 9 teeth would have a gear ratio of 4:3.

5 If the handles of a wheelbarrow are 1.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 220 lbs. load concentrated at a point 0.5 ft. from the axle?

52% Answer Correctly

	0
	73.3
	-64
	88.6

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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{220 \times 0.5}{1.5} \)
\( F_e = \frac{110.0}{1.5} \)
\( F_e = 73.3 \)