Answer
See the graph below.
Work Step by Step
We know, from Newton's second law, that
$$\sum F_x=ma_x$$
And in this case, there is only one force $F_x$ exerted on a 2-kg object.
$$ F_x=ma_x=2a_x$$
From the given figure, it is obvious that $a_x$ is not constant.
So, the force versus time graph is given by
- at $t=0$ s, $a_x=\bf 0$ m/s$^2$, and hence
$$F_{x }=2\cdot 0=\bf 0\;\rm N$$
$\color{red}{\rm(0\;s,0\;N)}$
- at $t=1$ s, $a_x=\bf 1.5$ m/s$^2$, and hence
$$F_{x }=2\cdot 1.5=\bf 3\;\rm N$$
$\color{red}{\rm(1\;s,3\;N)}$
- at $t=2$ s, $a_x=\bf 3$ m/s$^2$, and hence
$$F_{x }=2\cdot 3=\bf 6\;\rm N$$
$\color{red}{\rm(2\;s,6\;N)}$
- at $t=3$ s, $a_x=\bf 0$ m/s$^2$, and hence
$$F_{x }=2\cdot 0=\bf 0\;\rm N$$
$\color{red}{\rm(3\;s,0\;N)}$
- at $t=4$ s, $a_x=\bf 0$ m/s$^2$, and hence
$$F_{x }=2\cdot 0=\bf 0\;\rm N$$
$\color{red}{\rm(4\;s,0\;N)}$
Plugging all these dots into a force versus time graph, as you see below.