Answer
$v_2=10.7m/s$
Work Step by Step
According to the equation of motion
$\Sigma F_y=ma_y=0$
$\implies N+Fsin30-mg=0$
$\implies N+100sin30-20(9.81)=0$
$\implies N=146.2N$
Now the work of a constant force is given as
$\Sigma U_{1\rightarrow 2}=\Sigma F_x(s_2-s_1)$
$\implies \Sigma U_{1\rightarrow 2}=(Fcos30-\mu_k N)(s_2-s_1)$
$\implies \Sigma U_{1\rightarrow 2}=100cos30-(0.25\times 146.2)(25-15)=500.52N\cdot m$
We know that
$\frac{1}{2}mv^2_1+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv^2_2$
We plug in the known values to obtain:
$\frac{1}{2}20\times (8)^2+500.52=\frac{1}{2}\times 20\times v_2^2$
This simplifies to:
$v_2=10.67m/s\approx 10.7~m/s$
