- Mass of the object
- 80kg
- Speed at impact
- 10m/s
- Stopping distance
- 0.5m
8,000.00N
Open with these values8,000.00N
Result: 8,000.00 NAn impact spreads the kinetic energy over the stopping distance, so the average force is ½mv² divided by that distance. An 80 kg body at 10 m/s stopped over half a metre averages 8000 N; give it a full metre and the force halves. This is the average, not the peak.
8,000.00N
Open with these values200,000.00N
Open with these values4,375.00N
Open with these valuesF = ½ × m × v² ÷ d
| Mass, speed, distance | What it is | Average force |
|---|---|---|
| 50, 2, 0.1 | A 50 kg load stopped in 10 cm | 1000 |
| 70, 5, 0.2 | A person landing stiff-legged | 4375 |
| 80, 10, 0.5 | A body at 36 km/h, half a metre of give | 8000 |
| 1000, 20, 1 | A tonne at 72 km/h, one metre of crush | 200000 |
| 1500, 15, 0.8 | A 1.5 t car at 54 km/h | 210937.5 |
Divide the kinetic energy by the stopping distance: F = ½ × m × v² ÷ d. For an 80 kg body at 10 m/s stopped over 0.5 m the energy is 4000 J, so the average force is 8000 N.
The same energy has to be absorbed over a longer path, and force is energy divided by distance — doubling the distance halves the force. Crumple zones, airbags, climbing ropes and bending your knees all work by lengthening that path.
The average, assuming constant deceleration over the distance you enter. Real collisions are not constant, and the instantaneous peak can be several times higher. Use the figure to compare scenarios, not as the maximum at any single instant.
Ignoring air resistance, the speed just before landing is v = √(2 × g × h). A 1.5 m fall gives about 5.4 m/s. Enter that speed together with the mass and how far you sink on landing.
Divide the force by the object's own weight, that is by m × g. For 8000 N on an 80 kg body that is 8000 ÷ (80 × 9.81) ≈ 10 g. Textbooks round the 9.80665 standard value to 9.81 for this.
Information, not professional advice.
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