A set of crash tests consists of running a test car moving at a speed of 11.6 m/

Question

A set of crash tests consists of running a test car moving at a speed of 11.6 m/s (25.5 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 59.0 kg (129.8 lbs) dummy is found to move a distance of 0.840 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.

Using the direction of motion as positive direction, calculate the average acceleration of the dummy during that time (in g's) (use 1g=9.8 m/s2).

In a different car, the distance the dummy moves while being stopped is reduced from 0.840 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 = 10m/sec; Weight of dummy = 65kg and the distance traveled by the dummy after impact is S = 0.63m:

aS = -2Vx^2

0.63a = -2(10/2)^2 = -100/2

a = -79.365 ft/sec^2

using g = 9.817 m/sec^2

F = -ma = (W/g)a = -(65/9.817) 79.365 = -525.489 kg

For the second set of data where Vi = 10m/sec; W =l 65kg and S = 0.210m:

a = -238.095 m/sec^2
F = -1576.468 kg

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