## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $F = 6750~N$ (b) $F = 1.35\times 10^6~N$
(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 as; $F = ma$ $F = (60~kg)(112.5~m/s^2)$ $F = 6750~N$ (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 as; $F = ma$ $F = (60~kg)(22500~m/s^2)$ $F = 1.35\times 10^6~N$