A set of crash tests consists of running a test car moving at a speed of 11.8 m/

Question

A set of crash tests consists of running a test car moving at a speed of 11.8 m/s (26.4 mi/hr) into a solid wall. Strapped securely in an advanced seat beit system, a 71.0 kg (157 lbs) dummy is found to move a distance of 0.720 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time. The force that acts on the dummy has to do (negative) work on the dummy. It must slow the dummy and stop it. Thus the amounts of work equals the kinetic energy of the dummy. From the equation of work done by a force we can calculate that force. Using the direction of motion as the positive direction, calculate the average acceleration of the dummy during that time (in g s, with 1g = 9.81 m/s^2). In a different car, the distance the dummy moves while being stopped is reduced from 0.720 m to 0.210 m, calculate the average force on the dummy as that car stops.

Explanation / Answer

In Classical Newtonian Physics, The change in energy from one velocity to another is given by:

E = (1/2)m(Vf^2 - Vi^2)

Where;
E = change in energy
m = mass
Vf = Final Velocity
Vi = Initial velocity

But; E = FS

and

F = ma

Where;
F = the force that causes the decceleration
S = the distance traveled
a = the deceleration

Hence;

E = FS = maS = (1/2)m(Vf^2 -Vi^2)

aS = (Vf^2 - Vi^2)/2 ; or

2aS = Vf^2 - Vi^2

For the case at hand, the final velocity Vf is zero hence;

mas = -(1/2)mVi^2

The negative sign simply means that energy is being dissipated ( positive if being absorbed)

If instead of using Vf and Vi we use Vx where Vx is the average velocity, then;

Vx = (Vf + Vi)/2 =( 0 + Vi)/2 = Vi/2

Vi = 2Vx

maS = -(1/2)m(2Vx)^2 = -2mVx^2

Thertefore;

aS = -2Vx^2

For the given set of data where Vi =11.8m/sec;

Weight of dummy = 71kg and the distance traveled by the dummy after impact is S = 0.72m:

aS = -2Vx^2

0.72a = -2(11.8/2)^2 =

a = -96.69 ft/sec^2

using g = 9.817 m/sec^2

F = -ma = (W/g)a = -(71/9.817) 96.69 = -699.79 kg

For the second set of data where Vi = 11.8m/sec; W =l 71kg and S = 0.210m:

a = --96.69m/sec^2

F = -1576.468 kg

(The negative sign means that an inertial force is exerted on the dummy opposite its direction of motion).

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