Answer
At t = 1 s:
$F = 6.0~N$
At t = 4 s:
$F = 3.0~N$
At t = 7 s:
$F = 0$
Work Step by Step
The acceleration $a_x$ is equal to the slope of the $v_x$ versus time graph.
At t = 1 s:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{6~m/s-0}{2~s-0}$
$a_x = 3~m/s^2$
We can use $a_x$ to find the net force.
$F = ma_x$
$F = (2.0~kg)(3~m/s^2)$
$F = 6.0~N$
At t = 4 s:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{12~m/s-6~m/s}{6~s-2~s}$
$a_x = 1.5~m/s^2$
We can use $a_x$ to find the net force.
$F = ma_x$
$F = (2.0~kg)(1.5~m/s^2)$
$F = 3.0~N$
At t = 7 s:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{12~m/s-12~m/s}{8~s-6~s}$
$a_x = 0$
We can use $a_x$ to find the net force.
$F = ma_x$
$F = (2.0~kg)(0)$
$F = 0$
