SOLUTION (A)
Step 2: The sum of forces in y is equal
\(N-500\sin\theta-w=0\)
\(N=500\left(\frac{3}{5}\right)+98.1N\)\(N=398.1N\)
Step 3: The sum of forces in x is equal
\(500\cos\theta-300N=ma\)
\(400N-300N=ma\)
\(10\frac{m}{s^2}=a\)
\(v=vi+at\)
Step 4: Discarding the initial velocity because it is 0
\(v=20\frac{m}{s}\)
SOLUTION (B)
Step 1:  \(\Sigma Fy\)  The sums in forces in y are equal
\(N=ma\)
\(N=98.1N\)
Step 2: \(\Sigma Fx\)  The sums in forces in x are equal
\(F=\left(20t\right)N-N\)
\(F\left(t=4\right)=\left(20\left(4\right)N\right)-nF=79N\)