Question

A car moving at 70 km/h rams into an immobile steel wall. Its front is compressed by 0, 94 m. What average force must a seat belt exert in order to restrain a 75 kg passenger?

Answer 1

Speed before ram - $v_1=70 km/h$

Speed after ram (car stopped) $v_2=0 km/h$

The distance of carteh during ram $S=0.94m$

The distance during the ram:

$$S=v_1t+at^2/2$$

Where

$a -$ acceleration

$t -$ time during the ram

The Speed in the end:

$$v_2=v_1+at \implies t=(v_2-v_1)/a$$

Using this exprecion for the distance equation:

$$S=v_1(v_2-v_1)/a+(v_2-v_1)^2/(2a)$$

$$\implies$$

$$a=v_1(v_2-v_1)/S+(v_2-v_1)^2/(2S)$$

According Newton's 2nd law:

$$F=ma$$

Where m =75 kg, mass of the passenger

Using expresion for acceleration,

Using numbers for SI:

$$F=-195.5 kN$$

According to the Newton's 3rd law:

$$F_x=-F=195.5 kN$$

The direction of the force for belt is the same with direction of movement.

Answer: 195.5 kN