Verified Solution Science Work, Energy, and Simple Machines

A person uses an inclined plane to push a heavy box to a height of 3 m. The length of the inclined plane is 5 m. If the force required to push the box up the incline is 100 N, and the weight of the box is 300 N, calculate the efficiency of the inclined plane.

1 views 1 helpful Updated Jul 5, 2026
Solution ✔ Verified
  • A60%
  • B75%
  • C80%
  • D50%
Explanation

Actual Mechanical Advantage (AMA) = Load / Effort = 300 N / 100 N = 3. Velocity Ratio (VR) = Length of inclined plane / Vertical height = 5 m / 3 m = 1.67 (or 5/3). Efficiency = (AMA / VR) * 100% = (3 / (5/3)) * 100% = (9/5) * 100% = 1.8 * 100% = 80%. Wait, calculation error. (3 / (5/3)) = 3 * 3/5 = 9/5 = 1.8. This cannot be. AMA should be less than VR. Efficiency = (Work Output / Work Input) * 100%. Work Output = Load * Height = 300 N * 3 m = 900 J. Work Input = Effort * Length = 100 N * 5 m = 500 J. Efficiency = (900 J / 500 J) * 100% = 180%. This is incorrect for a real machine. Let's re-evaluate. The question implies the given force (100 N) is the actual effort required. So AMA is 300/100=3. IMA = VR = length/height = 5/3 = 1.66. Efficiency = AMA/IMA is not how it is defined if AMA is greater than IMA. The question is constructed such that AMA > VR which implies efficiency > 100% or there's an issue with the numerical values provided. Let's reconsider standard definition of IMA and AMA for inclined plane. IMA = L/H = 5/3 = 1.67. AMA = Load/Effort = 300/100 = 3. This means my calculated AMA (3) is greater than IMA (1.67), which is impossible for a real machine. There must be a typo in the question's numbers. If the *actual* effort required to lift a 300N box up a 5m ramp to 3m height is 100N, then the AMA = 3. But IMA for an inclined plane is Length/Height = 5/3 = 1.67. The AMA cannot be greater than IMA. This question would lead to an efficiency > 100%.

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