Four electric hoists in a maintenance facility lift cargo vertically at constant speeds. A… — Armed Services Vocational Aptitude Battery (ASVAB)

Four electric hoists in a maintenance facility lift cargo vertically at constant speeds. Arrange the hoists in order of their average mechanical power output, from lowest power to highest power.

1A hoist exerting a force of $500\text{ N}$ to lift a crate $4\text{ m}$ in $5\text{ s}$
2A hoist exerting a force of $250\text{ N}$ to lift a crate $10\text{ m}$ in $5\text{ s}$
3A hoist exerting a force of $300\text{ N}$ to lift a crate $6\text{ m}$ in $3\text{ s}$
4A hoist exerting a force of $800\text{ N}$ to lift a crate $3\text{ m}$ in $3\text{ s}$

Answer

The correct sequence from lowest to highest mechanical power is the hoist delivering

400\text{ W}

(

500\text{ N} \times 4\text{ m} / 5\text{ s}

), followed by the hoist delivering

500\text{ W}

(

250\text{ N} \times 10\text{ m} / 5\text{ s}

), then the hoist delivering

600\text{ W}

(

300\text{ N} \times 6\text{ m} / 3\text{ s}

), and ending with the hoist delivering

800\text{ W}

(

800\text{ N} \times 3\text{ m} / 3\text{ s}

Mechanical power equals the rate of doing work (

P = \frac{F \times d}{t}

). Computing the power output yields

400\text{ W}

for the

500\text{ N}

load over

4\text{ m}

5\text{ s}

500\text{ W}

for the

250\text{ N}

load over

10\text{ m}

5\text{ s}

600\text{ W}

for the

300\text{ N}

load over

6\text{ m}

3\text{ s}

, and

800\text{ W}

for the

800\text{ N}

load over

3\text{ m}

3\text{ s}

. This produces the ascending sequence

400\text{ W} \rightarrow 500\text{ W} \rightarrow 600\text{ W} \rightarrow 800\text{ W}

Step-by-Step Solution

Recall the formula for mechanical power in terms of force, distance, and time.

P = \frac{W}{t} = \frac{F \times d}{t}

where power (

P

) is in Watts (

\text{W}

), force (

F

) in Newtons (

\text{N}

), distance (

d

) in meters (

\text{m}

), and time (

t

) in seconds (

\text{s}

Power measures the rate at which work is performed.

Calculate the power output for each hoist.

First hoist:

(300\text{ N} \times 6\text{ m}) / 3\text{ s} = 1800\text{ J} / 3\text{ s} = 600\text{ W}

. Second hoist:

(500\text{ N} \times 4\text{ m}) / 5\text{ s} = 2000\text{ J} / 5\text{ s} = 400\text{ W}

. Third hoist:

(800\text{ N} \times 3\text{ m}) / 3\text{ s} = 2400\text{ J} / 3\text{ s} = 800\text{ W}

. Fourth hoist:

(250\text{ N} \times 10\text{ m}) / 5\text{ s} = 2500\text{ J} / 5\text{ s} = 500\text{ W}

Determines the exact numerical power for all four mechanical scenarios.

Arrange the resulting power values in ascending order.

400\text{ W} < 500\text{ W} < 600\text{ W} < 800\text{ W}

Satisfies the requirement to order from lowest power to highest power.

Key Concept

Power is the rate of doing work, defined as work divided by time (

P = \frac{W}{t} = \frac{F \times d}{t}

), and is measured in Watts.

Estimated Time:1m 15s

Question

Answer

Step-by-Step Solution

Key Concept

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