#### Answer

(a) W = 120 J
(b) W = -108 J
(c) 24.2 J

#### Work Step by Step

(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 $79.6^{\circ}$
$W = F~d~cos(\theta)$
$W = (30.0~N)(4.47~m)~cos(79.6^{\circ})$
$W = 24.2~J$