Simple machine performance: A 500 kgf load is lifted through 13 cm by an effort of 25 kgf moving 650 cm. What is the efficiency of the lifting machine (in %)?

Introduction / Context:

Efficiency compares useful output to input in any simple machine. For lifting devices, output work equals load times load displacement, whereas input work equals effort times effort displacement. Expressing all quantities in consistent force units (e.g., kgf) and distance units (e.g., cm) keeps the ratio dimensionless and directly convertible to percent.

Given Data / Assumptions:

• Load = 500 kgf.
• Load is lifted by 13 cm.
• Effort = 25 kgf moves through 650 cm.
• Neglect elastic energy storage and other losses beyond what efficiency captures.

Concept / Approach:

Efficiency η = Output work / Input work. For vertical lifting at constant speed, Output = Load * Load distance; Input = Effort * Effort distance. No unit conversion is needed if the same force and length units are used in numerator and denominator.

Step-by-Step Solution:

Verification / Alternative check:

Mechanical advantage MA = Load / Effort = 500 / 25 = 20. Velocity ratio VR = Effort distance / Load distance = 650 / 13 = 50. Efficiency = MA / VR = 20 / 50 = 0.40 → 40%, confirming the result.

Why Other Options Are Wrong:

• 30%, 50%, 55%, 60% do not match the computed ratio of works.

Common Pitfalls:

• Mixing units (e.g., metres with centimetres) between numerator and denominator.
• Using load distance and effort distance in the wrong places.

Final Answer:

40%