1. The problem involves a mass of 16 kg, an angle of 60 degrees, a distance of 10 m, and a fraction 2/5. We need to understand what to solve, likely a physics problem involving forces or work. 2. Assuming this is a problem about calculating the component of a force or weight along a slope or direction, we use the formula for the component of weight along an incline: $$ F = mg \sin(\theta) $$ where $m$ is mass, $g$ is acceleration due to gravity (9.8 m/s²), and $\theta$ is the angle. 3. Calculate the weight force: $$ W = mg = 16 \times 9.8 = 156.8 \text{ N} $$ 4. Calculate the component along the 60° angle: $$ F = 156.8 \times \sin(60^\circ) = 156.8 \times \frac{\sqrt{3}}{2} \approx 156.8 \times 0.866 = 135.7 \text{ N} $$ 5. The fraction 2/5 might represent a coefficient or ratio; if it is a coefficient of friction or similar, multiply: $$ F_{effective} = 135.7 \times \frac{2}{5} = 135.7 \times 0.4 = 54.28 \text{ N} $$ 6. If the distance is 10 m and we want work done by this force: $$ W = F_{effective} \times d = 54.28 \times 10 = 542.8 \text{ J} $$ Final answer: The work done or effective force component is approximately 542.8 J (joules).