Ideal machine with large velocity ratio — an ideal machine has mechanical advantage (MA) = 100. If the load moves 2 m, how far must the effort move?

Introduction / Context:
For an ideal (lossless) simple machine, mechanical advantage equals velocity ratio. This means the distance the effort travels is amplified relative to the load movement by the same factor that the effort reduces the necessary force. Understanding this equivalence is central to hoists, jacks, and block-and-tackle systems.

Given Data / Assumptions:

• Mechanical advantage MA = 100 (ideal, so efficiency = 100%).
• Load displacement s_load = 2 m.
• Velocity ratio VR = distance moved by effort / distance moved by load.

Concept / Approach:

In an ideal machine, MA = VR. Therefore the kinematic requirement is that the effort must move VR times the distance of the load. Multiply the given load displacement by MA to obtain the effort displacement.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

A power balance for an ideal machine gives Effort * s_effort = Load * s_load; with Load/Effort = MA = 100, s_effort/s_load = 100, consistent.

Why Other Options Are Wrong:

20 m reflects MA = 10; 2.5 m and 2 m ignore the high velocity ratio; 0.02 m reverses the relation.

Common Pitfalls (misconceptions, mistakes):

Confusing MA with efficiency; forgetting that in ideal machines the gain in force is paid for by a proportional loss in distance moved.

Final Answer:

200 m