Impact Force
Impact forces acting on falling objects hitting the ground, cars crashing and similar cases.
The dynamic kinetic energy of a moving object, like a falling ball or a driving car, can be expressed as
E = 1/2 m v ^{ 2 } (1)
where
E = kinetic (dynamic) energy (J, ft lb)
m = mass of the object (kg, slugs )
v = velocity of the object (m/s, ft/s)
In an impact  like a car crash  the work made by the impact force slowing down an moving object over a distance by deforming the crumple zone can be expressed as
W = F _{ avg } s (2)
where
W = work done (J, ft lb)
F _{ avg } = average impact force during deformation (N, lb _{ f } )
s = deformation distance, crumple zone (m, ft)
When a crumple zone deforms in a car crash the average impact force is designed to be as constant as possible.
In an impact where the object is not deformed  the work made by the impact force slowing down the moving object equals to the work done by a spring force  and can be expressed as
W = 1/2 F _{ max } s
= 1/2 k s ^{ 2 } (2b)
where
W = work done (J, ft lb)
F _{ max } = maximum force at the end of the deformation (N, lb _{ f } )
k = spring constant
s = deformation distance (m, ft)
In a car crash the dynamic energy is converted to work and equation 1 and 2 can be combined to
F _{ avg } s = 1/2 m v ^{ 2 } (3)
The average impact force can be calculated as
F _{ avg } = 1/2 m v ^{ 2 } / s (3b)
The deformation slowdown distance can be calculated as
s = 1/2 m v ^{ 2 } / F _{ avg } (3c)
Note!  The deformation slowdown distance is very important and the key to limit the forces acting on passengers in a car crash.
Example  Car Crash
A car with a mass of 2000 kg drives with speed 60 km/h (16.7 m/s) before it crashes into a massive concrete wall. The front of the car impacts 0.5 m (the deformation distance).
The impact force can be calculated as
F _{ max } = 1/2 (2000 kg) (16.7 m/s) ^{ 2 } / (0.5 m)
= 558 kN
Note that the gravitation force (weight) acting on the car is only
F _{ w } = m g
= (2000 kg) (9.81 m/s ^{ 2 } )
= 19.6 kN
The impact creates a force 28 times gravity!!
A person sitting inside the car with seat belts on will deaccelerate with a force 28 times gravity . Note that the National Highway Traffic Safety Administration (NHTSA) states that "the maximum chest acceleration shall not exceed 60 times gravity for time periods longer than 3 milliseconds ". For a car crash with 90 km/h (25 m/s) the deacceleration will be 64 times gravity (same parameters as above).
 60 mph = 96.6 km/h
Impact Force from a Falling Object
The dynamic energy in a falling object at the impact moment when it hits the ground can be calculated as
E = F _{ weight } h
= m a _{ g } h (4)
where
F _{ weight } = force due to gravity  or weight (N, lb _{ f } )
a _{ g } = acceleration of gravity (9.81 m/s ^{ 2 } , 32.17405 ft/s ^{ 2 } )
h = falling height (m)
If the dynamic energy from the fall is converted to impact work  equation 2 and 4 can be combined to
F _{ avg } s = m a _{ g } h (5)
The impact force can be expressed as
F _{ avg } = m a _{ g } h / s (5b)
The deformation slowdown distance can be expressed as
s = m a _{ g } h / F _{ avg } (5c)
Example  a Falling Car
The same car as above falls from height 14.2 m and crashes on the crumple zone with the front down on a massive concrete tarmac. The front impacts 0.5 m (slow down distance) as above. The impact force can be calculated as
F _{ avg } = (2000 kg) (9.81 m/s ^{ 2 } ) (14.2 m) / (0.5 m)
= 558 kN
Note!  a car crash in 90 km/h (25 m/s) compares to a fall from 32 m !!
Example  a Person falling from a Table
A person with weight (gravitational force) of 200 lbs (lb _{ f } ) falls from a 4 feet high table.
The energy of the falling body when it hits the ground can be calculated using (4) as
E = (200 lb _{ f } ) (4 ft)
= 800 ft lb
The impact on a human body can be difficult to determine since it depends on how the body hits the ground  which part of the body, the angle of the body and/or if hands are used to protect the body and so on.
For this example we use an impact distance of 3/4 inch (0.0625 ft) to calculate the impact force:
F _{ avg } = (800 ft lb) / (0.0625 ft)
= 12800 lb _{ f }
In metric units  person with weight 90 kg, falling distance 1.2 m and impact distance 2 cm :
E = (90 kg) (9.81 m/s ^{ 2 } ) (1.2 m)
= 1059 J
F _{ avg } = (1059 J) / (0.02 m)
= 53 kN
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