Answer
At t = 1 s:
$F = 8.0~N$
At t = 4 s:
$F = 0$
At t = 7 s:
$F = -12~N$
Work Step by Step
The acceleration $a_x$ is equal to the slope of the $v_x$ versus time graph.
At t = 1 s:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{12~m/s-0}{3.0~s-0}$
$a_x = 4.0~m/s^2$
We can use $a_x$ to find the net force:
$F = ma_x$
$F = (2.0~kg)(4.0~m/s^2)$
$F = 8.0~N$
At t = 4 s:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{12~m/s-12~m/s}{6.0~s-3.0~s}$
$a_x = 0$
We can use $a_x$ to find the net force;
$F = ma_x$
$F = (2.0~kg)(0)$
$F = 0$
At t = 7 s:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{0-12~m/s}{8.0~s-6.0~s}$
$a_x = -6.0~m/s^2$
We can use $a_x$ to find the net force;
$F = ma_x$
$F = (2.0~kg)(-6.0~m/s^2)$
$F = -12~N$
