#### Answer

(a) $F = 6750~N$
(b) $F = 1.35\times 10^6~N$
(c) seat belt and air bag:
The force is 11.5 times the person's weight.
no seat belt and no air bag:
The force is 2300 times the person's weight.

#### Work Step by Step

(a) We can find the rate of deceleration as:
$a = \frac{0-v_0^2}{2x}$
$a = \frac{-(15~m/s)^2}{(2)(1~m)}$
$a = -112.5~m/s^2$
We can use the magnitude of deceleration to find the net force on the person.
$F = ma$
$F = (60~kg)(112.5~m/s^2)$
$F = 6750~N$
(b) We can find the rate of deceleration as:
$a = \frac{0-v_0^2}{2x}$
$a = \frac{-(15~m/s)^2}{(2)(0.005~m)}$
$a = -22500~m/s^2$
We can use the magnitude of deceleration to find the net force on the person.
$F = ma$
$F = (60~kg)(22500~m/s^2)$
$F = 1.35\times 10^6~N$
(c) We can find the person's weight;
$weight = mg = (60~kg)(9.80~m/s^2)$
$weight = 588~N$
When the seat belt and air bag stops the person:
$F = \frac{6750~N}{588~N} = 11.5$
The force is 11.5 times the person's weight.
When the person is not restrained:
$F = \frac{1.35\times 10^6~N}{588~N} = 2300$
The force is 2300 times the person's weight.