A mechanic needs to move a 150\text{ kg} engine block onto a storage shelf that is 120\tex… — Armed Services Vocational Aptitude Battery (ASVAB)

A mechanic needs to move a $150\text{ kg}$ engine block onto a storage shelf that is $120\text{ cm}$ above the floor. He uses a frictionless inclined plane and applies a constant effort force of $300\text{ N}$ parallel to the ramp's surface to push the block at a constant speed. Assuming the acceleration due to gravity is $10\text{ m/s}^2$ , what is the required length of the inclined plane, and how does the total work done using the ramp compare to the work required to lift the engine vertically?

$6.0\text{ m}$ ; total work done is equal to lifting it verticallyAnswer
B
$6.0\text{ m}$ ; total work done is less than lifting it vertically
C
$0.6\text{ m}$ ; total work done is less than lifting it vertically
D
$600\text{ m}$ ; total work done is equal to lifting it vertically

Answer

The required length is

6.0\text{ m}

, and the total work done is equal to the work required to lift the engine vertically.

The weight of the engine block is

1500\text{ N}

. To lift it

1.2\text{ m}

vertically requires

1800\text{ J}

of work. Using a frictionless ramp, the exact same

1800\text{ J}

of work is performed by applying a smaller

300\text{ N}

force over a longer distance (

6.0\text{ m}

). Simple machines change the ratio of force and distance but do not alter the total work done.

Step-by-Step Solution

Calculate the weight of the engine block and convert the height to meters.

Weight =

150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}

. Height =

1.2\text{ m}

Standard SI units (meters) and forces (Newtons) are required for accurate work and energy calculations.

Apply the principle of conservation of work (Work Input = Work Output) to solve for the length.

300\text{ N} \times L = 1500\text{ N} \times 1.2\text{ m} \rightarrow 300L = 1800 \rightarrow L = 6.0\text{ m}

In a frictionless system, the input work done by the effort force over the ramp's length equals the output work needed to lift the object vertically.

Compare the total work done in both scenarios.

Work on ramp (

300\text{ N} \times 6.0\text{ m} = 1800\text{ J}

) equals Work lifted vertically (

1500\text{ N} \times 1.2\text{ m} = 1800\text{ J}

Simple machines like inclined planes provide a mechanical advantage by spreading the same amount of work over a longer distance, reducing the required force.

Key Concept

Ideal Mechanical Advantage and Work Conservation on an Inclined Plane

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