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Physics
Inclined plane — Raising a load
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Drag the ramp foot to change its length. Raise the load and compare force and work with a vertical lift.
Model force gauge
50 N0–550 N
Distance pulled: 0 / 4 mWork so far: 0 J
Steady upward pull; startup and stopping omitted.Arrows show direction
Drag the foot outward, then select Pull steadily.
μ = 0 ignores friction. Other values are assumptions about the contacting surfaces.
Changing the setup starts a new pull; the pinned result is kept.
Force for steady pulling
50 N
Vertical lift: 200 N
- Distance pulled
- 0 m
- Height gained
- 0 m
Work so far
0 J
- ● Potential energy gained
- 0 J
- ▧ Due to friction
- 0 J
Ready to pull
Less pulling force, but a longer path. Run the trial to follow the work along the ramp.
Reach the top to pin a result, then change the ramp and try again.
Why this result?
Input work = pulling force × distance
A = F × s = mgz + f × s
F = mg(h + μ√(L² − h²)) / L
Trip efficiency: 100%
Friction converts this part to internal energy of the load and ramp.
Model notes
- The load slides steadily up a straight ramp with a parallel pulling force. An ideal puller controls the motion; startup, stopping and static friction are omitted. Pause freezes the demonstration, not a static force balance.
- We use g ≈ 10 m/s². The ramp, load and rope are idealized; air drag, deformation and puller mass are omitted. Gauge and ruler readings are calculated from the model.
- The kinetic friction coefficient is a constant assumption, not a particular material. Input work equals gravitational potential energy gained plus internal energy generated by friction in the contact system, not disappearing energy or the load's temperature.
- Only upward travel is modeled. Acceleration, downward sliding, rolling, tipping and the transition onto the platform are excluded. The crate size is illustrative; ramp geometry and rise use the same scale.