#### Answer

(a) F = -12 N
(b) F = 12 N

#### Work Step by Step

$x = (2t^3-3t^2)~m$
$v = \frac{dx}{dt} = (6t^2-6t)~m/s$
$a = \frac{dv}{dt} = (12~t-6)~m/s^2$
(a) At t = 0:
$a = [(12)(0) - 6]~m/s^2$
$a = -6~m/s^2$
We can use the acceleration to find the net horizontal force $F$
$F = ma$
$F = (2.0~kg)(-6~m/s^2)$
$F = -12~N$
(b) At t = 1 s:
$a = [(12)(1) - 6]~m/s^2$
$a = 6~m/s^2$
We can use the acceleration to find the net horizontal force $F$
$F = ma$
$F = (2.0~kg)(6~m/s^2)$
$F = 12~N$