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Difficulty: MediumSimple Machines

An inclined plane of length 10 m10\text{ m} is used to raise a heavy crate of load 900 N900\text{ N} to a height of 2 m2\text{ m}. If an effort force of 250 N250\text{ N} is applied parallel to the inclined surface to push the crate up at constant speed, what is the efficiency of the simple machine?

  1. $72\%Answer
  2. B
    $139\%
  3. C
    $36\%
  4. D
    $5.6\%

Answer

The efficiency of the inclined plane is 72%72\%.
The mechanical advantage of the machine is MA=900 N250 N=3.6MA = \frac{900\text{ N}}{250\text{ N}} = 3.6, and its velocity ratio is VR=10 m2 m=5VR = \frac{10\text{ m}}{2\text{ m}} = 5. Dividing MAMA by VRVR and multiplying by 100%100\% gives an efficiency of 72%72\%.

Step-by-Step Solution

1
Calculate the Mechanical Advantage (MA)
MA=LoadEffort=900 N250 N=3.6MA = \frac{\text{Load}}{\text{Effort}} = \frac{900\text{ N}}{250\text{ N}} = 3.6
Mechanical advantage is defined as the ratio of load force to effort force.
2
Calculate the Velocity Ratio (VR) of the inclined plane
VR=Length of planeHeight of plane=10 m2 m=5VR = \frac{\text{Length of plane}}{\text{Height of plane}} = \frac{10\text{ m}}{2\text{ m}} = 5
Velocity ratio for an inclined plane is the ratio of distance moved by effort along the incline to distance moved by load vertically.
3
CalculatetheEfficiency(η)Calculate the Efficiency (\eta)
\eta = \left(\frac{MA}{VR}\right) \times 100\% = \left(\frac{3.6}{5}\right) \times 100\% = 72\%
Efficiency is the ratio of Mechanical Advantage to Velocity Ratio, expressed as a percentage.

Key Concept

Efficiency of an Inclined Plane
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Topics

Simple Machines

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