WBBSE Madhyamik Physical Science study materials
WBBSE Madhyamik Physical Science study resources

Ramp Loading, Friction and Efficiency Trade-off

WBBSE Madhyamik Physical Science · Put your knowledge to work in context.

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Ramp Loading, Friction and Efficiency Trade-off

Workers are loading 50 kg sacks onto a truck using a wooden ramp 4.0 m long that is raised 1.2 m at the truck end. One worker pushes a sack up the ramp with a steady force of 180 N parallel to the ramp; he takes 6.0 s for one sack. He claims that the ramp reduces the total work needed to lift a sack, because the force he applies is much less than the sack's weight. Another worker suggests replacing the ramp with a shorter 2.0 m ramp of the same height to save space. The friction force between sack and ramp is found to be 30 N and is assumed to stay the same on the shorter ramp. Take g = 10 m/s². The workers want to know whether the long ramp is actually better and how the change would affect the effort needed and the efficiency.

1. Calculate the useful work, the work done by the worker, and the efficiency of the 4.0 m ramp. Show that the given 180 N effort is consistent with a 30 N friction force.

Read the worked answer

Useful work = mgh = 50×10×1.2 = 600 J. Work input = F×s = 180×4.0 = 720 J. Efficiency = 600/720×100 ≈ 83.3%. The extra work is 720−600 = 120 J, used against friction: f = 120/4.0 = 30 N, consistent with the given friction.

2. Assess the worker's claim that the ramp reduces the total work needed.

Read the worked answer

The ramp does not reduce the total work; it increases it from the ideal 600 J to 720 J because friction does extra work. What the ramp reduces is the force needed at any instant: 180 N instead of 500 N. The worker's statement confuses force with work.

3. For the shorter 2.0 m ramp, calculate the effort needed if friction is still 30 N and the efficiency. Decide which ramp is better for a worker who can push with at most 250 N, and comment on the efficiency trade-off.

Read the worked answer

On the 2.0 m ramp, the component of weight along the ramp is mg sinθ = 500×(1.2/2.0) = 300 N. Adding 30 N friction gives effort = 330 N. Work input = 330×2.0 = 660 J; efficiency = 600/660×100 ≈ 90.9%. The shorter ramp is more efficient but needs 330 N, above the worker's 250 N limit. For that worker the long ramp is necessary; the shorter ramp would suit someone stronger.