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Difficulty: HardInclined Planes

A heavy transport truck with a mass of 10 metric tons10\text{ metric tons} (1 metric ton=1000 kg1\text{ metric ton} = 1000\text{ kg}) needs to deliver supplies to a mountain facility located 500 m500\text{ m} vertically above its current position. A direct, steep road up the mountainside to the facility is 1000 m1000\text{ m} long. Instead, the truck takes a winding switchback road that is 4000 m4000\text{ m} long to reach the exact same facility. Assuming an ideal system with no friction and using g=10 m/s2g = 10\text{ m/s}^2, what is the required effort force on the switchback road, and how does the total work done compare to using the direct road?

  1. A
    The effort force is 12.5 N12.5\text{ N}, and the total work done is the same as the direct road.
  2. B
    The effort force is 12,500 N12,500\text{ N}, and the total work done is one-fourth of the work required on the direct road.
  3. The effort force is 12,500 N12,500\text{ N}, and the total work done is the same as the direct road.Answer
  4. D
    The effort force is 1,250 N1,250\text{ N}, and the total work done is the same as the direct road.

Answer

The effort force is 12,500 N12,500\text{ N}, and the total work done is the same as the direct road.
The correct answer accurately calculates the truck's weight (100,000 N100,000\text{ N}), finds the ideal mechanical advantage of the switchback road (4000 m/500 m=84000\text{ m} / 500\text{ m} = 8), and divides the weight by the IMA to find the effort force (12,500 N12,500\text{ N}). It also correctly identifies that in an ideal system, the total work done depends only on the vertical height and load, remaining constant regardless of the path taken.

Step-by-Step Solution

1
Calculate the total weight (load force) of the truck.
Weight = 10,000 kg×10 m/s2=100,000 N10,000\text{ kg} \times 10\text{ m/s}^2 = 100,000\text{ N}.
The mass must be converted to standard SI units (kilograms) and multiplied by gravity to find the downward force required to lift it.
2
Calculate the Ideal Mechanical Advantage (IMA) of the switchback road.
IMA=LengthHeight=4000 m500 m=8\text{IMA} = \frac{\text{Length}}{\text{Height}} = \frac{4000\text{ m}}{500\text{ m}} = 8.
The mechanical advantage of an inclined plane is the ratio of its slope length to its vertical height.
3
Calculate the required effort force on the switchback road.
Effort Force=LoadIMA=100,000 N8=12,500 N\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{100,000\text{ N}}{8} = 12,500\text{ N}.
In an ideal system, the required effort force is equal to the load divided by the mechanical advantage.
4
Compare the total work done between the two paths.
The total work remains the same for both the direct road and the switchback road.
According to the conservation of energy, an inclined plane reduces the effort force by increasing the distance traveled, but the total work done (Force ×\times Distance) remains constant in an ideal system.

Key Concept

Ideal Mechanical Advantage of Inclined Planes and Conservation of Work
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Topics

Inclined Planes

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