Average force of a 0.5 kg hammer at 10 m/s stopping in 5 mm

Answer: average impact force = 5,000 N (in kilonewtons 5 kN, as multiples of the object's weight 1,019.7 × weight).

What we know

  • Mass: 0.5 kg
  • Impact speed: 10 m/s
  • Stopping distance: 5 mm = 0.005 m (converted to the unit the formula takes)

Step by step

Given
m = 0.5 kg, v = 10 m/s, d = 0.005 m
Formula
F = m·v² / (2·d)
Substitute
F = 0.5 × 10² / (2 × 0.005)
Result
F = 5,000 N

All results

Average impact force5,000 N
In kilonewtons5 kN
As multiples of the object's weight1,019.7 × weight

Open the impact force calculator with these numbers → and change any of them.

The physics

When a moving object is brought to rest, its kinetic energy (½mv²) is absorbed over the stopping distance. Work equals force times distance, so the average force is the energy divided by that distance: F = ½mv² / d. The shorter the stop, the larger the force — which is the whole reason crumple zones, crash mats and packaging exist.

Standard acceleration of gravity →