1. **State the problem:** We need to find the efficiency of an electric motor that uses 78.0 J of electrical energy to accelerate a 280 g toy car from rest to 14.0 m/s over 8.3 m in 3.80 s. 2. **Formula for efficiency:** Efficiency $\eta$ is given by the ratio of useful output energy to input energy: $$\eta = \frac{\text{useful energy output}}{\text{energy input}} \times 100\%$$ 3. **Identify useful energy output:** The useful energy output is the kinetic energy gained by the toy car: $$KE = \frac{1}{2} m v^2$$ where $m$ is mass and $v$ is final velocity. 4. **Convert mass to kg:** $$m = 280\ \text{g} = 0.280\ \text{kg}$$ 5. **Calculate kinetic energy:** $$KE = \frac{1}{2} \times 0.280 \times (14.0)^2 = 0.14 \times 196 = 27.44\ \text{J}$$ 6. **Calculate efficiency:** $$\eta = \frac{27.44}{78.0} \times 100\% = 35.18\%$$ **Final answer:** The efficiency of the motor is approximately **35.2%**.