Answer
Applying Newton's second law to the $x$ (directed uphill) and $y$ (normal to the
inclined plane) axes, we obtain
$$
\begin{array}{c}{F-m g \sin \theta=m a} \\ {F_{N}-m g \cos \theta=0}\end{array}
$$
The second equation allows us to solve for the angle the inclined plane makes with the
horizontal:
$$
\theta=\cos ^{-1}\left(\frac{F_{N}}{m g}\right)=\cos ^{-1}\left(\frac{13.41 \mathrm{N}}{(4.00 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}\right)=70.0^{\circ}
$$
From the equation for the $x$ -axis, we find the acceleration of the block to be
$$
\begin{aligned} a=\frac{F}{m}-g \sin \theta=& \frac{50 \mathrm{N}}{4.00 \mathrm{kg}}-\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \sin 70.0^{\circ}=3.29 \mathrm{m} / \mathrm{s}^{2} \\ \\ \text { Using the kinematic equation } v^{2} &=v_{0}^{2}+2 a d, \text { the speed of the block when } \\ d=3.00 \mathrm{m} \text { is } \\ v=\sqrt{2 a d} &=\sqrt{2\left(3.29 \mathrm{m} / \mathrm{s}^{2}\right)(3.00 \mathrm{m})}=4.44 \mathrm{m} / \mathrm{s} \end{aligned}
$$
Work Step by Step
Applying Newton's second law to the $x$ (directed uphill) and $y$ (normal to the
inclined plane) axes, we obtain
$$
\begin{array}{c}{F-m g \sin \theta=m a} \\ {F_{N}-m g \cos \theta=0}\end{array}
$$
The second equation allows us to solve for the angle the inclined plane makes with the
horizontal:
$$
\theta=\cos ^{-1}\left(\frac{F_{N}}{m g}\right)=\cos ^{-1}\left(\frac{13.41 \mathrm{N}}{(4.00 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}\right)=70.0^{\circ}
$$
From the equation for the $x$ -axis, we find the acceleration of the block to be
$$
\begin{aligned} a=\frac{F}{m}-g \sin \theta=& \frac{50 \mathrm{N}}{4.00 \mathrm{kg}}-\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \sin 70.0^{\circ}=3.29 \mathrm{m} / \mathrm{s}^{2} \\ \\ \text { Using the kinematic equation } v^{2} &=v_{0}^{2}+2 a d, \text { the speed of the block when } \\ d=3.00 \mathrm{m} \text { is } \\ v=\sqrt{2 a d} &=\sqrt{2\left(3.29 \mathrm{m} / \mathrm{s}^{2}\right)(3.00 \mathrm{m})}=4.44 \mathrm{m} / \mathrm{s} \end{aligned}
$$
