Answer
The work done on the particle by the force is $~~12.0~J$
Work Step by Step
$F = ma$
We can use the acceleration versus position graph to draw a force versus position graph. Note that the basic shape of both graphs will be similar and the scale of the vertical axis on the force versus position graph is $F_s = (2.00~kg)(6.0~m/s^2) = 12.0~N$
To find the work done, we can calculate the area under the force versus position curve.
From $x=0$ to $x=9.0~m$, this area can be divided into six parts, including a triangle (0 to 1.0 m), a rectangle (1.0 m to 4.0 m), a triangle (4.0 m to 5.0 m), a triangle (5.0 m to 6.0 m), a rectangle (6.0 m to 8.0 m), and a triangle (8.0 m to 9.0 m).
We can find each area separately:
$A_1 = \frac{1}{2}(12.0~N)(1.0~m) = 6.0~J$
$A_2 = (12.0~N)(3.0~m) = 36.0~J$
$A_3 = \frac{1}{2}(12.0~N)(1.0~m) = 6.0~J$
$A_4 = \frac{1}{2}(-12.0~N)(1.0~m) = -6.0~J$
$A_5 = (-12.0~N)(6.0~m) = -24.0~J$
$A_6 = \frac{1}{2}(-12.0~N)(1.0~m) = -6.0~J$
We can find the work done by the force:
$W = 6.0~J+36.0~J+6.0~J-6.0~J-24.0~J-6.0~J = 12.0~J$
The work done on the particle by the force is $~~12.0~J$.
