Calculate G-Force & Deceleration for 4500kg Object Moving @ 5kph

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SUMMARY

The discussion centers on calculating G-force and deceleration for a 4500kg object traveling at 5kph (1.38m/s) that stops over a distance of 15cm (0.15m). The calculated deceleration is 6.348m/s², derived from the formula a = V²/2L. The impact force is determined to be 28566N, and the pressure exerted is approximately 1941.9kg/m² or 19.044N/cm². The calculations are confirmed to be accurate, providing essential data for the user's professional argument.

PREREQUISITES
  • Understanding of basic physics concepts, specifically Newton's laws of motion.
  • Familiarity with kinematic equations, particularly L = V²/2a.
  • Knowledge of pressure calculations, including the formula P = F/A.
  • Ability to convert units between metric and imperial systems, such as kg to N and cm² to m².
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  • Research "Newton's laws of motion" for a deeper understanding of force and acceleration.
  • Study "kinematic equations" to master calculations involving velocity, distance, and acceleration.
  • Explore "pressure calculations in physics" to understand how force is distributed over an area.
  • Learn about "unit conversions in physics" to effectively switch between different measurement systems.
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This discussion is beneficial for engineers, physicists, safety professionals, and anyone involved in accident analysis or impact force calculations.

exalto1
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Hello...

If an object weighs 4500kg and travels @ 5kph (138cm/s) then stops in 15cm:

what is Gforce?

Decceleration?

Also, if object was stopped via a collision and if contact face is 100cmx15cm then what is impact force kg/sq.cm?

This information is really important to me for a professional argument that may save my job so ANY help/resources would be immensely appreciated.

Yes I did have an accident at work... :redface:

My only defence is going to have to be with cold science.

At 36, my algebra has rusted through as this is the first time I've had to find "x" so really need some whizzy help.

Thanks to anyone kind enough to help.

David
 
Last edited:
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If something travels with the initial velocity V...
1.38m/s

and then stops over the distance L,.....
0.15m

then, assuming constant decelleration,

L = V^2/2a.........

0.15m=1.38m^2/2 x 6.348
0.15=1.9044/12.696


where V - initial velocity
a - absolute value of decceleration.
V^2 means "V squared"

From this, you can find

a = V^2/2L......... ...

a=1.38m^2/0.30m
a=6.348m/s

From the Newton's law,

a = F/M.......... ...

6.348m/s=28566n/4500kg

where F is the force that acted during the slowdown and M is the mass.

We then have

F = MV^2/2L.......

4500kg x 1.38m/s^2 /2 x 0.15m
4500 x 1.9044 / 0.30m
8569.8/0.30
F=28566n

Use M in kilograms, V in meters per second and L in meters. You'll get the force in Newtons.

I don't know what is Gforce.

Then the pressure is by definition the force divided by area. Diveide the force you've obtained by the area and you'll get pressure in Newtons per unit of area (whatever units you use)...

28566n/1.5m=19044n/sq m
or 1941.9kg/sq m

If someone could check this please I'd be grateful as I'm a little woozy with painkillers.

Thank you so very much.
 
Hi exalto1,

exalto1 said:
Then the pressure is by definition the force divided by area. Diveide the force you've obtained by the area and you'll get pressure in Newtons per unit of area (whatever units you use)...[/B]
28566n/1.5m=19044n/sq m
or 1941.9kg/sq m

I'm not sure about these numbers. The area in square meters is .15 m^2.

G-force is usually the acceleration measured in units of g=9.8 m/s^2.
 
Hello alphysicist...

Are these figures better:P = 28566N/1500cm^2=19.044N/cm^2 or
P =~1.94kg/cm^2 or
P=~19400kg/m^2
 
exalto1 said:
Hello alphysicist...

Are these figures better:


P = 28566N/1500cm^2=19.044N/cm^2 or
P =~1.94kg/cm^2 or
P=~19400kg/m^2

Your math look okay to me, if those are the units you want. Using kg/cm^2 for the pressure reminded me that a unit that some people use for force is the gram-force. I would think that's probably what they meant by Gforce, so your choice of mass/area as the units for pressure would be what they wanted.
 
Thank you so very much Al
 

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