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

(a) The angle $\theta$ between the force vector and the displacement vector is $37.0^{\circ}$ $W = F~d~cos(\theta)$ $W = (30.0~N)(5.00~m)~cos(37.0^{\circ})$ $W = 120~J$ (b) The angle $\theta$ between the force vector and the displacement vector is $127.0^{\circ}$ $W = F~d~cos(\theta)$ $W = (30.0~N)(6.00~m)~cos(127.0^{\circ})$ $W = -108~J$ (c) We can find the magnitude of the displacement. $d = \sqrt{(-2.00~m)^2+(4.00~m)^2}$ $d = 4.47~m$ We can find the angle $\alpha$ above the (-x)-axis. $tan(\alpha) = \frac{4.00~m}{2.00~m}$ $\alpha = arctan(\frac{4.00~m}{2.00~m})$ $\alpha = 63.4^{\circ}$ The angle $\theta$ between the force vector and the displacement vector is $180.0^{\circ} - 63.4^{\circ}-37.0^{\circ}$, which is $79.6^{\circ}$ $W = F~d~cos(\theta)$ $W = (30.0~N)(4.47~m)~cos(79.6^{\circ})$ $W = 24.2~J$