Examkin

Question

Difficulty: MediumInclined Planes

A flight line technician must load a cargo pallet with a mass of 240 kg240\text{ kg} into a transport aircraft. The aircraft's cargo deck is 1.5 meters1.5\text{ meters} above the tarmac. To assist with loading, the technician deploys a straight, 6.0-meter6.0\text{-meter} long rigid ramp equipped with frictionless rollers. Assuming the acceleration due to gravity is 10 m/s210\text{ m/s}^2, which of the following identifies the ideal effort force required to push the pallet up the ramp, and correctly compares the total work performed to lifting the pallet vertically?

  1. 600 N600\text{ N}; the total work performed is the same as lifting the pallet vertically.Answer
  2. B
    600 N600\text{ N}; the total work performed is less than lifting the pallet vertically.
  3. C
    60 N60\text{ N}; the total work performed is the same as lifting the pallet vertically.
  4. D
    60 N60\text{ N}; the total work performed is less than lifting the pallet vertically.

Answer

The effort force is 600 N600\text{ N}, and the total work performed is exactly the same as lifting the pallet vertically.
The correct answer accurately calculates both the required force and applies the principle of work conservation. First, weight is determined by multiplying mass by gravity (240 kg×10 m/s2=2400 N240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}). The ideal mechanical advantage (IMA) of the ramp is its length divided by its height (6.0 m/1.5 m=46.0\text{ m} / 1.5\text{ m} = 4). The effort force is the weight divided by the IMA (2400 N/4=600 N2400\text{ N} / 4 = 600\text{ N}). Finally, while the ramp reduces the effort force, it requires moving the object over a longer distance, meaning the total work performed (600 N×6.0 m=3600 Joules600\text{ N} \times 6.0\text{ m} = 3600\text{ Joules}) is identical to lifting it straight up (2400 N×1.5 m=3600 Joules2400\text{ N} \times 1.5\text{ m} = 3600\text{ Joules}).

Step-by-Step Solution

1
Calculate the weight (load force) of the cargo pallet.
W=240 kg×10 m/s2=2400 NW = 240\text{ kg} \times 10\text{ m/s}^2 = 2400\text{ N}
Mass must be converted to weight (force) before mechanical advantage formulas can be properly applied.
2
Determine the Ideal Mechanical Advantage (IMA) of the inclined plane.
IMA=6.0 m1.5 m=4\text{IMA} = \frac{6.0\text{ m}}{1.5\text{ m}} = 4
IMA for an inclined plane is the ratio of the effort distance (ramp length) to the load distance (vertical height).
3
Calculate the ideal effort force.
Effort Force=2400 N4=600 N\text{Effort Force} = \frac{2400\text{ N}}{4} = 600\text{ N}
The effort force is equal to the total load force divided by the mechanical advantage of the machine.
4
Evaluate the total work performed in both scenarios.
Work via ramp: 600 N×6.0 m=3600 J600\text{ N} \times 6.0\text{ m} = 3600\text{ J}. Work vertical: 2400 N×1.5 m=3600 J2400\text{ N} \times 1.5\text{ m} = 3600\text{ J}. Work is equal.
According to the principle of conservation of energy, simple machines trade increased distance for reduced effort force, leaving the total work unchanged.

Key Concept

Mechanical advantage reduces effort force by increasing distance, but total work is conserved.
Rate this question

Topics

Inclined Planes

More questions from this topic