Answer
(a) $F_{net} = 11 ~N$
(b) Between 2.0 and 6.0 seconds, the net force is zero.
(c) $F_{net} = 2.75 ~N$
Work Step by Step
(a) The maximum net force occurs between 0 and 2.0 seconds when the velocity graph has the steepest slope. The slope is $(8.0~m/s) / (2.0~s)$, which is $a = 4.0~m/s^2$.
$F_{net} = ma = (2.75~kg)(4.0~m/s^2) = 11~N$
(b) Between 2.0 and 6.0 seconds, the acceleration is zero, so the net force is zero.
(c) At t = 8.5 seconds, the slope of the graph is $(8.5~m/s) / 8.5~s$ so the acceleration is $a = 1.0~m/s^2$.
$F_{net} = ma = (2.75~kg)(1.0~m/s^2) = 2.75 ~N$
