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