Let's break this down **step by step**. --- ### **Step 1: Understand what happens from x = 0.180 m to x = −0.240 m** - **Initial position:** \( x_i = 0.180\,\mathrm{m} \) - **Final position:** \( x_f = -0.240\,\mathrm{m} \) - **Change in potential energy:** \( \Delta U = U_f - U_i = -1.01 \times 10^{-16}\,\mathrm{J} \) - Negative means the electron *loses* potential energy (or gains kinetic energy). --- ### **Step 2: Use energy conservation to find the speed at x = −0.240 m** **Energy Conservation:** \[ \Delta K + \Delta U = 0 \implies K_f - K_i = -\Delta U \] \[ K_f = K_i - \Delta U \] Kinetic energies: \[ K = \frac{1}{2} m v^2 \] Let \( v_i \) be the speed at \( x = 0.180\,\mathrm{m} \), and \( v_f \) be the speed at \( x = -0.240\,\mathrm{m} \). So, \[ \frac{1}{2} m v_f^2 = \frac{1}{2} m v_i^2 - \Delta U \] \[ v_f^2 = v_i^2 - \frac{2}{m}\Delta U \] --- ### **Step 3: Plug in known values** #### **First, find \( v_i \)** From previous part: - At \( x = -2.00\,\mathrm{cm} = -0.0200\,\mathrm{m} \): initial speed \( v_0 \). - At \( x = 0.180\,\mathrm{m} \): speed is \( v_1 = \frac{1}{2} v_0 \). - Change in potential energy from \( x = -0.0200\,\mathrm{m} \) to \( x = 0.180\,\mathrm{m} \) is \( \Delta U_1 = 4.81 \times 10^{-17}\,\mathrm{J} \). Apply conservation of energy for this leg: \[ \frac{1}{2} m v_1^2 = \frac{1}{2} m v_0^2 - \Delta U_1 \] \[ v_1^2 = v_0^2 - \frac{2}{m}\Delta U_1 \] But \( v_1 = \frac{1}{2} v_0 \): \[ \left(\frac{1}{2} v_0\right)^2 = v_0^2 - \frac{2}{m}\Delta U_1 \] \[ \frac{1}{4} v_0^2 = v_0^2 - \frac{2}{m}\Delta U_1 \] \[ \frac{2}{m}\Delta U_1 = v_0^2 - \frac{1}{4} v_0^2 = \frac{3}{4} v_0^2 \] \[ v_0^2 = \frac{4}{3} \cdot \frac{2}{m} \Delta U_1 = \frac{8 \Delta U_1}{3m} \] So, \[ v_1 = \frac{1}{2} v_0 = \frac{1}{2} \sqrt{\frac{8 \Delta U_1}{3m}} \] #### **Now, plug \( v_i = v_1 \) into the formula for \( v_f \):** \[ v_f^2 = v_1^2 - \frac{2}{m}\Delta U_2 \] Where \( \Delta U_2 = -1.01 \times 10^{-16}\,\mathrm{J} \) (**from \( x=0.180 \) to \( -0.240 \)**). So, \[ v_f^2 = v_1^2 - \frac{2}{m}(-1.01 \times 10^{-16}) = v_1^2 + \frac{2}{m}(1.01 \times 10^{-16}) \] Now substitute \( v_1^2 = \frac{1}{4} v_0^2 = \frac{1}{4} \cdot \frac{8\Delta U_1}{3m} = \frac{2\Delta U_1}{3m} \): \[ v_f^2 = \frac{2\Delta U_1}{3m} + \frac{2}{m}(1.01 \times 10^{-16}) \] \[ v_f^2 = \frac{2}{m} \left( \frac{\Delta U_1}{3} + 1.01 \times 10^{-16} \right ) \] --- #### **Plug in the numbers:** - \( \Delta U_1 = 4.81 \times 10^{-17}\,\mathrm{J} \) - \( m = 9.11 \times 10^{-31}\,\mathrm{kg} \) (mass of electron) - \( \Delta U_2 = -1.01 \times 10^{-16}\,\mathrm{J} \) \[ v_f^2 = \frac{2}{9.11 \times 10^{-31}}\left(\frac{4.81 \times 10^{-17}}{3} + 1.01 \times 10^{-16} \right ) \] Calculate \( \frac{4.81 \times 10^{-17}}{3} \approx 1.603 \times 10^{-17} \): \[ v_f^2 = \frac{2}{9.11 \times 10^{-31}}(1.603 \times 10^{-17} + 1.01 \times 10^{-16}) \] \[ = \frac{2}{9.11 \times 10^{-31}}(1.1703 \times 10^{-16}) \] \[ = \frac{2 \times 1.1703 \times 10^{-16}}{9.11 \times 10^{-31}} \] \[ = \frac{2.3406 \times 10^{-16}}{9.11 \times 10^{-31}} \] \[ = 2.570 \times 10^{14} \] \[ v_f = \sqrt{2.570 \times 10^{14}} \approx 1.60 \times 10^7\,\mathrm{m/s} \] --- ## **Summary of Steps:** **Step 1:** Wrote the energy conservation equation for the electron as it moves from \( x = 0.180\,\mathrm{m} \) to \( x = -0.240\,\mathrm{m} \), relating changes in kinetic and potential energy. **Step 2:** Expressed the final speed in terms of the initial speed at \( x = 0.180\,\mathrm{m} \) and the given potential energy change, using prior results for \( v_1 \). **Step 3:** Plugged in the numbers, using the electron mass and the given changes in potential energy, to solve for the final speed. --- ## **Final Answer:** \[ \boxed{v_f = 1.60 \times 10^7\,\mathrm{m/s}} \] This is the speed of the electron at \( x = -0.240\,\mathrm{m} \).