Efficiency of a simple lifting machine (work output / work input): A 50 kg load is raised by 2.5 cm while an effort of 12.5 kg moves through 40 cm. Assuming ideal alignment and no other losses than friction, what is the machine efficiency?

Introduction / Context:
Efficiency of simple machines compares useful output work to input work. Expressing loads and efforts in consistent “kgf-cm” (or converting to newton-metres consistently) leads to the same efficiency ratio, provided the same gravity factor multiplies both sides and cancels out.

Given Data / Assumptions:

• Load L = 50 kgf (kilogram-force), rises by h_L = 2.5 cm.
• Effort E = 12.5 kgf, moves by h_E = 40 cm.
• Quasi-static motion; ignore kinetic energy change.

Concept / Approach:
Work output = L * h_L; work input = E * h_E (when using kgf and centimetres). Efficiency η = (Output / Input) * 100%. Alternatively, use mechanical advantage MA = L / E and velocity ratio VR = h_E / h_L; then η = (MA / VR) * 100%. Both approaches are equivalent if units are consistent.

Step-by-Step Solution:

Verification / Alternative check:
The two independent routes (MA/VR and direct work) agree. Using newton-metres would multiply both numerator and denominator by the same g factor, cancelling out and giving the same 25%.

Why Other Options Are Wrong:

• 20%, 40%, 50%, 60%: arise from arithmetic errors or mixing distances/units.

Common Pitfalls:
Confusing cm and m; mixing kg (mass) and kgf (force) without consistency; using MA alone as “efficiency”.

Final Answer:
25%