The Overloaded Minibus on the Salang Pass
A minibus carrying 14 passengers climbs the Salang Pass. Its total mass is 2,800 kg. The engine delivers a constant forward force of 6,000 N. Along the climb, a combined resistive force from friction, air drag, and gravity along the slope acts backward at 4,200 N. The road rises 1 meter for every 12 meters traveled. The driver keeps the speed constant at 15 m/s for a 600 m stretch. A passenger notes the engine is working hard but the speedometer needle does not move. Later, on a flat section, the same engine force acts against only 1,200 N of resistance.
1. On the climb, is the minibus accelerating? Use the forces to justify your answer, and compute the net force.
Read the worked answer
No acceleration. Constant speed means the net force is zero.
Net force = forward force − backward forces = 6,000 N − 4,200 N = 1,800 N.
This is a contradiction unless we read the numbers carefully: the stated resistive force (4,200 N) already includes the component of gravity along the slope, so the actual net force along the road must be zero for constant velocity. The 1,800 N surplus means the given 4,200 N cannot be the total resistance during steady climbing; either the engine force is lower or resistance is higher. Judgement point: a constant-speed claim and unbalanced forces cannot both be true. The physically consistent reading is that total backward force equals 6,000 N on the climb.
2. Estimate the power the engine must deliver on the climb if total resistance is actually 6,000 N at 15 m/s, and compare it with the flat section.
Read the worked answer
Power = force × speed.
Climb: P = 6,000 N × 15 m/s = 90,000 W = 90 kW.
Flat: net driving force needed to hold constant speed equals resistance = 1,200 N, so P = 1,200 × 15 = 18,000 W = 18 kW.
The climb demands about 5 times the power of the flat section. This explains why the engine 'works hard' while speed stays fixed: extra power goes into overcoming gravity along the slope, not into speeding up.
3. A passenger claims that because the road rises 1 m per 12 m, the vehicle gains 1 m of height every 12 m of travel. If it covers 600 m of road, how much height does it gain, and does this match the 'constant speed' energy picture?
Read the worked answer
Height gained = 600 m ÷ 12 = 50 m.
Gravitational potential energy gained = mgh = 2,800 × 9.8 × 50 ≈ 1,372,000 J ≈ 1.37 MJ.
At 15 m/s, the 600 m takes 40 s. The power needed just to lift the vehicle = 1.37 MJ ÷ 40 s ≈ 34 kW. This is a large share of the 90 kW climb power, confirming gravity along the slope is a major backward force. Since kinetic energy does not change (constant speed), all engine work above the friction/drag share goes into potential energy - consistent with a steady climb.