One movable pulley: force and rope travel

A pulley block with 1 movable pulley lifts a 10 kg load at constant speed; 2 rope segments support the load. The load rises 2 m. Find the minimum pulling force and the distance the rope end moves. (Take g = 10 N/kg.)

Step-by-step

1. Weight G = mg = 10×10 = 100 N
   Find the weight of the object being lifted first
2. F = G/n = 100/2 = 50 N
   n rope segments share the load, so each needs only G/n (n = 2 ✓)
3. s = n·h = 2×2 = 4 m
   Less force means more distance: the rope end moves n times the lift height (work is not saved ✓)

Why it works

When n rope segments carry the load, each segment only needs the weight divided by n, but the rope end must be pulled n times as far — saving force always costs distance.

Common mistakes

• Answering 100 N and forgetting to divide by the number of rope segments.
• Thinking force and distance are both saved, and giving 1 m for the rope end.

Related

incline-friction vertical-throw pulley-one-movable-8kg-lift-1-2m