(a) Use the conservation of energy principle. The initial kinetic energy (KE) at A is \(\frac{1}{2} \times 0.2 \times 5^2\). The final KE at B is \(\frac{1}{2} \times 0.2 \times 3^2\). The change in potential energy (PE) is \(0.2gh\).

Equating initial and final energies:

\(\frac{1}{2} \times 0.2 \times 5^2 = 0.2gh + \frac{1}{2} \times 0.2 \times 3^2\)

Solving for \(h\):

\(h = 0.8\) m

(b) Apply the work-energy principle from A to C. The initial KE is \(\frac{1}{2} \times 0.2 \times 5^2\). The work done against resistance is 3.1 J. The change in PE is \(0.2g \times 0.5\). The final KE at C is \(\frac{1}{2} \times 0.2 \times v^2\).

Using the work-energy equation:

\(\frac{1}{2} \times 0.2 \times 5^2 - 3.1 + 0.2g \times 0.5 = \frac{1}{2} \times 0.2 \times v^2\)

Solving for \(v\):

\(Speed = 2 m/s\)