## University Physics with Modern Physics (14th Edition)

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.