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

\(F=\left(20t\right)N-N\)

\(F\left(t=4\right)=\left(20\left(4\right)N\right)-nF=79N\)