Reducing Force: Understanding Energy Absorption in Crashes

[Miraj Kayastha]

Does reducing acceleration mean reducing force acting upon the object. And does this mean absorbing energy from the object?

If so where is the energy absorbed in case of crashes where air bags inflate to reduce force?

 

[Drakkith]

Does reducing acceleration mean reducing force acting upon the object. And does this mean absorbing energy from the object?

Yes, reducing the force reduces the acceleration. It doesn't require that energy be absorbed. For example, lifting the accelerator up a little reduces the acceleration of my car, but no energy is absorbed anywhere. Instead it's that I'm not spending as much energy in the first place.

If so where is the energy absorbed in case of crashes where air bags inflate to reduce force?

The air bag itself absorbs the energy. Note that an air bag works because it increases the time for the person's head to decelerate to a stop. Impacting the steering wheel or the dashboard would bring your head to a stop MUCH faster. Since acceleration is dv/dt (d means delta, which means a change in the value of v and t), reducing the time means that acceleration is higher. For example, going from 10 m/s to 0 m/s over 10 seconds would be: 10/10 = 1 m/s².
Going from 10 m/s to 0 m/s over 1 second would be: 10/1 = 10 m/s².

Since the equation for force is F=MA, a higher acceleration means a higher force.

With or without the airbag the energy absorbed is still the same.