ASVAB Mechanical Comprehension Practice Test 944813 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.82
Score	0%	56%

Review

What is the first step to solving a problem where multiple forces are acting on an object?

61% Answer Correctly

	calculate the net force
	calculate the total force
	calculate potential energy
	calculate kinetic energy

Solution

In mechanics, multiple forces are often acting on a particular object and, taken together, produce the net force acting on that object. Like force, net force is a vector quantity in that it has magnitude and direction.

2 A 330 lb. barrel is rolled up a 10 ft. ramp to a platform that's 3 ft. tall. What effort is required to move the barrel?

53% Answer Correctly

	102 lbs.
	297 lbs.
	99 lbs.
	49.5 lbs.

Solution

This problem describes an inclined plane and, for an inclined plane, 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 10 ft. = 330 lbs. x 3 ft.
F_e = \( \frac{990 ft⋅lb}{10 ft.} \) = 99 lbs.

3 If the green arrow in this diagram represents 200 ft⋅lb of work, how far will the box move if it weighs 50 pounds?

72% Answer Correctly

	16 ft.
	25 ft.
	12 ft.
	4 ft.

Solution

The Law of Work states that the work put into a machine is equal to the work received from the machine under ideal conditions. In equation form, that's:

W_in = W_out
F_effort x d_effort = F_resistance x d_resistance

In this problem, the effort work is 200 ft⋅lb and the resistance force is 50 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
200 ft⋅lb = 50 lbs. x d_resistance
d_resistance = \( \frac{200ft⋅lb}{50 lbs.} \) = 4 ft.

4 What is the power output of a 2 hp engine that's 65% efficient?

39% Answer Correctly

	0 \( \frac{ft⋅lb}{s} \)
	715 \( \frac{ft⋅lb}{s} \)
	178.8 \( \frac{ft⋅lb}{s} \)
	130 \( \frac{ft⋅lb}{s} \)

Solution

\( Efficiency = \frac{Power_{out}}{Power_{in}} \times 100 \)
Solving for power out: \( P_{o} = \frac{E \times P_{i}}{100} \)
Knowing that 1 hp = 550 \( \frac{ft⋅lb}{s} \), P_i becomes 2 hp x 550 \( \frac{ft⋅lb}{s} \) = 1100 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{65 \times 1100 \frac{ft⋅lb}{s}}{100} \) \( = \frac{71500 \frac{ft⋅lb}{s}}{100} \) = 715 \( \frac{ft⋅lb}{s} \)

For any given surface, the coefficient of static friction is ___________ the coefficient of kinetic friction.

54% Answer Correctly

	lower than
	opposite
	higher than
	equal to

Solution

For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).