#### Answer

The work done by the force on the grocery cart is -155 J.

#### Work Step by Step

We can find the magnitude of the force.
$F = \sqrt{(30~N)^2+(-40~N)^2}= 50~N$
We can find the angle below the +x-axis.
$tan(\theta) = \frac{40~N}{30~N}$
$\theta = arctan(\frac{40~N}{30~N}) = 53^{\circ}$
We can find the magnitude of the displacement.
$s = \sqrt{(-9.0~m)^2+(-3.0~m)^2}= 9.5~m$
We can find the angle below the (-x)-axis.
$tan(\theta) = \frac{3.0~m}{9.0~m}$
$\theta = arctan(\frac{3.0~m}{9.0~m}) = 18^{\circ}$
The angle between the displacement vector and the force vector is $180^{\circ} - 53^{\circ}-18^{\circ}=109^{\circ}$
We can find the work done by the force.
$W = F~s~cos(\theta)$
$W = (50~N)(9.5~m)~cos(109^{\circ})$
$W = -155~J$
The work done by the force on the grocery cart is -155 J.