Answer
(a) $a = 2.50\times 10^{14}~m/s^2$
(b) $t = 1.20\times 10^{-8}~s$
(c) $F_{net} = 2.28\times 10^{-16}~N$
Work Step by Step
(a) $a = \frac{v^2-v_0^2}{2x}$
$a = \frac{(3.00\times 10^6~m/s)^2-0}{(2)(0.0180~m)}$
$a = 2.50\times 10^{14}~m/s^2$
(b) $t = \frac{v-v_0}{a} = \frac{(3.00\times 10^6~m/s)-0}{2.50\times 10^{14}~m/s^2}$
$t = 1.20\times 10^{-8}~s$
(c) $F_{net} = ma$
$F_{net} = (9.11\times 10^{-31}~kg)(2.50\times 10^{14}~m/s^2)$
$F_{net} = 2.28\times 10^{-16}~N$
