91Ó°ÊÓ

The force, $\mathrm{F}=12\mathrm{N}$

The displacement, $\stackrel{→}{\mathrm{d}}=\left(2.00\hat{\mathrm{i}}-4.00\hat{\mathrm{j}}+30.0\hat{\mathrm{k}}\right)$

The work-energy theorem states that the net work done by forces on an object is equal to the change in its kinetic energy.

Using the work-kinetic energy theorem, you have

$\begin{array}{rcl}\mathbf{∆}\mathbf{K}& \mathbf{=}& \mathbf{W}\\ & \mathbf{=}& \stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{d}}\\ & \mathbf{=}& \mathbf{¹ó»å³¦´Ç²õÏ•}\end{array}$

Calculate the magnitude of the displacement as below.

$\begin{array}{rcl}\mathrm{d}& =& \sqrt{{\left(2.00\mathrm{m}\right)}^2+{\left(-4.00\mathrm{m}\right)}^2+{\left(3.00\mathrm{m}\right)}^2}\\ & =& \sqrt{\left(4+16+9\right)}\mathrm{m}\\ & =& \sqrt{29}\mathrm{m}\\ & =& 5.39\mathrm{m}\end{array}$

The angle between the force and the displacement if the change in the particle’s kinetic energy is,

$∆\mathrm{K}=+30.0\mathrm{J}$

Therefore, the angle is,

$\begin{array}{rcl}\mathrm{ϕ}& =& {\cos }^{-1}\left(\frac{∆\mathrm{K}}{\mathrm{Fd}}\right)\\ & =& {\cos }^{-1}\left(\frac{30.0\mathrm{J}}{\left(12.0\mathrm{N}\right)\left(5.39\mathrm{m}\right)}\right)\\ & =& {\cos }^{-1}\left(0.464\right)\\ & =& 62.3^{°}\end{array}$

Hence, the angle between the force and the displacement is $62.3^{°}$.

The angle between the force and the displacement if the change in the particle’s kinetic energy is,

$∆\mathrm{K}=-30.0\mathrm{J}$

Therefore, the angle is,

$\begin{array}{rcl}\mathrm{ϕ}& =& {\cos }^{-1}\left(\frac{∆\mathrm{K}}{\mathrm{Fd}}\right)\\ & =& {\cos }^{-1}\left(\frac{-30.0\mathrm{J}}{\left(12.0\mathrm{N}\right)\left(5.39\mathrm{m}\right)}\right)\\ & =& {\cos }^{-1}\left(-0.464\right)\\ & =& 117.{64}^{°}\\ \mathrm{ϕ}& ≈& {118}^{°}\end{array}$

Hence, the angle between the force and the displacement is ${118}^{°}$.