Answer
$v_{f} = 5.244\frac{m}{s}$
Work Step by Step
Split the angled forces into their x and y components and summate them to find the acceleration of the block. Use the relationship between acceleration and displacement to obtain the velocity of the block.
The external force is angled. Therefore, split it into x and y components using triangular ratios:
y-component of the external force $= 500(\frac{3}{5}) =300$ Newtons
x-component of the external force $= 500(\frac{4}{5}) =400$ Newtons
In addition, the block is experiencing a gravitational force $((10 kg)(9.8\frac{m}{s^{2}})$ in the figure), a spring force ($F_{s}$ in the figure), and a normal force ($F_{N}$ in the figure). Summate the x and y components:
(1) $\sum F_{x}=ma_{x}=400-F_{s} = 400-(500)(s)$
(2) $\sum F_{y}=ma_{y}=F_{N}-300-(10)(9.81)$
By definition, $vdv=ads$. Apply this definition to equation (1) to obtain the velocity:
$a_{x}=\frac{400-500s}{10} = 40-50s$
vdv$=(40-50s)ds$
$\int_{0}^{v_{f}}vdv=\int_{0}^{0.5}(40-50s)ds$
$\frac{v_{f}^{2}}{2}-0=40s-25s^{2}|_{0}^{0.5}$
$\frac{v_{f}^{2}}{2} = 13.75$
$v_{f}^{2} = 27.5$
$v_{f} = 5.244\frac{m}{s}$
Turns out, equation (2) was not necessary.
