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Loading ramp: useful work, real work, and the efficiency

SPH3U Grade 11 Physics · Energy and Society

Two movers slide a 120 kg crate up a 3.60 m aluminum ramp onto a truck deck 75.0 cm above the parking lot, pulling with a steady 410 N directed along the ramp. The crate moves at constant speed the whole way up.

Given
  • m = 120 kg — Mass of the crate
  • L = 3.6 m — Length of the ramp
  • h = 75 cm — Height of the truck deck
  • F = 410 N — Pull along the ramp
Determine
  1. (a)the useful work done on the crate
  2. (b)the work the movers actually do
  3. (c)the efficiency of the ramp
  4. (d)the pull the same ramp would need if it were frictionless
Step 1 of 4(a) · solve for Potential energy

The useful output is only what gravity keeps: mgh through the 0.750 m of HEIGHT. Using the 3.60 m ramp length here is the classic wreck — the ramp is the path, but gravity only bills for the rise.

Rearranged for U
U=mghU = m g h
Your values, in your units
U=(120 kg) (9.80665 m/s2) (75 cm)U = \left( 120\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 75\ \text{cm} \right)
Converted to base units
U=(120 kg) (9.80665 m/s2) (0.75 m)U = \left( 120\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 0.75\ \text{m} \right)
Answer
U=882.6 JU = 882.6\ \text{J}

Carried onward at full precision, not this rounded figure.

Open the Gravitational Potential Energy (U = mgh) solver →

Step 2 of 4(b) · solve for Work

The input side: the pull acts along the ramp, so here the 3.60 m IS the distance that counts, with θ = 0° because force and motion line up. Each measurement gets used exactly once, on its own side of the ledger.

Rearranged for W
W=Fdcos⁡θW = F d \cos\theta
Your values, in your units
W=(410 N) (3.6 m) cos⁡(0 ∘)W = \left( 410\ \text{N} \right) \, \left( 3.6\ \text{m} \right) \, \cos \left( 0\ ^{\circ} \right)
Answer
W=1.476 kJW = 1.476\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Work (W = Fd cos θ) solver →

Step 3 of 4(c) · solve for Efficiency

Output over input, joules over joules — never a force over a force taken from mismatched directions. The 40% that goes missing is not mysterious: it left as heat where the crate's base scrubbed along the aluminum.

Rearranged for η
η=WoutWin\eta = \frac{W_{out}}{W_{in}}
882.6 Jcarried from step 1
1.476 kJcarried from step 2
Your values, in your units
η=(882.599 J)(1,476 J)\eta = \frac{\left( 882.599\ \text{J} \right)}{\left( 1{,}476\ \text{J} \right)}
Answer
η=597.97 per mille\eta = 597.97\ \text{per mille}

Carried onward at full precision, not this rounded figure.

Open the Machine Efficiency solver →

Step 4 of 4(d) · solve for Force

Run the ideal case: if the ramp charged nothing for friction, the pull would only need to supply mgh over 3.60 m — about 245 N. The gap up to the real 410 N is friction's share, read directly in newtons.

Rearranged for F
F=Wdcos⁡θF = \frac{W}{d \cos\theta}
882.6 Jcarried from step 1
Your values, in your units
F=(882.599 J)(3.6 m) cos⁡(0 ∘)F = \frac{\left( 882.599\ \text{J} \right)}{\left( 3.6\ \text{m} \right) \, \cos \left( 0\ ^{\circ} \right)}
Answer
F=245.17 NF = 245.17\ \text{N}

Carried onward at full precision, not this rounded figure.

Open the Work (W = Fd cos θ) solver →

Therefore lifting the crate onto the deck is worth 883 J, the movers actually spend 1476 J, the ramp runs at 60% efficiency — and a frictionless ramp would have asked for only 245 N of the 410 N pull.

Why this order

An efficiency is a fraction with two independently earned sides, and the chain builds them strictly apart: the output in part (a) uses only the mass and the height, the input in part (b) uses only the pull and the ramp length, and they first meet in part (c). The error the layout is designed to prevent is cross-contamination — mgh computed with the 3.60 m slope, or F·d computed with the 0.750 m rise. The discipline is dimensional and geometric at once: gravity is a vertical force, so it only does work through vertical distance; the pull is a along-ramp force, so it only does work through along-ramp distance. Two forces, two distances, no sharing.

Part (d) reframes the same 60% in the unit a mover can feel. The ideal pull, 245 N, is what the incline would charge purely for lifting — and it is exactly the real 410 N multiplied by the efficiency, because over the same 3.60 m the work ratio IS the force ratio. That 245 N is also mg sin θ in disguise (the ramp rises 0.750 in 3.60, and 1176.798 N × 0.75/3.6 = 245.17 N), which ties this chain back to the incline decomposition without ever measuring the angle. Notice finally what the ramp is for despite its losses: even at 60%, pulling 410 N beats dead-lifting the crate's full 1177 N. A simple machine that wastes 40% of the work can still be the difference between a job two people can do and one they cannot — efficiency and usefulness are different columns in the ledger.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.