# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 34

(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.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.