1. **State the problem:** Hockey Player A has a weight of 784 N and is moving at +5 m/s. Hockey Player B applies a force over 0.5 s to stop Player A completely. We need to find the average magnitude of this force. 2. **Identify known values:** - Weight of Player A, $W = 784$ N - Initial velocity, $v_i = +5$ m/s - Final velocity, $v_f = 0$ m/s (complete stop) - Time interval, $\Delta t = 0.5$ s 3. **Find mass of Player A:** Weight is related to mass by $W = mg$, where $g = 9.8$ m/s$^2$ (acceleration due to gravity). $$m = \frac{W}{g} = \frac{784}{9.8} = 80\text{ kg}$$ 4. **Calculate acceleration:** Using the formula for acceleration: $$a = \frac{v_f - v_i}{\Delta t} = \frac{0 - 5}{0.5} = -10\text{ m/s}^2$$ The negative sign indicates deceleration. 5. **Calculate force using Newton's second law:** $$F = ma = 80 \times (-10) = -800\text{ N}$$ 6. **Interpret the result:** The magnitude of the force is the absolute value: $$|F| = 800\text{ N}$$ **Final answer:** The average magnitude of the force applied by Hockey Player B is 800 N.