Let’s solve each problem in **two steps** with explanations and then provide the final answer. --- ## **6. Electron Speed from Kinetic Energy** **Given:** - Mass of electron, \( m = 9.11 \times 10^{-31} \) kg - Kinetic energy, \( KE = 6.7 \times 10^{-20} \) J ### **Step 1: Write the Kinetic Energy Formula** The kinetic energy of an object is given by: \[ KE = \frac{1}{2}mv^2 \] **Explanation:** Here, \( KE \) is the kinetic energy, \( m \) is the mass, and \( v \) is the speed. We need to solve for \( v \). --- ### **Step 2: Solve for the Speed \( v \)** Rearrange the equation to solve for \( v \): \[ v = \sqrt{\frac{2KE}{m}} \] Substitute the values: \[ v = \sqrt{\frac{2 \times 6.7 \times 10^{-20}}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{\frac{1.34 \times 10^{-19}}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{1.471 \times 10^{11}} \] \[ v \approx 3.83 \times 10^5 \ \text{m/s} \] **Explanation:** By plugging in the known values, we find the speed of the electron. --- ### **Final Answer for 6:** \[ \boxed{v \approx 3.8 \times 10^5 \ \text{m/s}} \] --- ## **7. Work Done by a Force** **Given:** - Force: \( \vec{F} = (2\, \text{N})\hat{i} + (3\, \text{N})\hat{j} \) - Initial position: \( \vec{r}_i = (2\, \text{m})\hat{i} + (3\, \text{m})\hat{j} \) - Final position: \( \vec{r}_f = (3\, \text{m})\hat{i} + (\, \text{m})\hat{j} \) --- ### **Step 1: Find the Displacement Vector \( \Delta \vec{r} \)** \[ \Delta \vec{r} = \vec{r}_f - \vec{r}_i = [(3-2)\hat{i} + (-3)\hat{j}] = (1\hat{i} - 3\hat{j})\, \text{m} \] **Explanation:** Displacement is simply the difference between the final and initial positions for each component. --- ### **Step 2: Calculate the Work Done \( W = \vec{F} \cdot \Delta \vec{r} \)** \[ W = (2\hat{i} + 3\hat{j}) \cdot (1\hat{i} - 3\hat{j}) \] \[ W = (2 \times 1) + (3 \times -3) = 2 - 9 = -7\, \text{J} \] **Explanation:** Work is the dot product of force and displacement vectors. Multiply corresponding components and add. --- ### **Final Answer for 7:** \[ \boxed{W = -7\, \text{J}} \]

# Step-by-Step Solutions ## **6. Electron Speed from Kinetic Energy** **Given:** - Mass of electron, \( m = 9.11 \times 10^{-31} \) kg - Kinetic energy, \( KE = 6.7 \times 10^{-20} \) J ### **Step 1: Write the Kinetic Energy Formula** The formula for kinetic energy is: \[ KE = \frac{1}{2}mv^2 \] **Explanation:** In this equation, \( KE \) represents kinetic energy, \( m \) is mass, and \( v \) is the speed of the electron. We need to rearrange this formula to find \( v \). --- ### **Step 2: Solve for the Speed \( v \)** Rearranging the kinetic energy formula gives: \[ v = \sqrt{\frac{2KE}{m}} \] Substituting the known values: \[ v = \sqrt{\frac{2 \times 6.7 \times 10^{-20}}{9.11 \times 10^{-31}}} = \sqrt{\frac{1.34 \times 10^{-19}}{9.11 \times 10^{-31}}} \] Calculating the square root: \[ v \approx \sqrt{1.471 \times 10^{11}} \approx 3.83 \times 10^5 \ \text{m/s} \] **Explanation:** By substituting the kinetic energy and mass values into the formula, we determine the speed of the electron to be approximately \( 3.83 \times 10^5 \ \text{m/s} \). --- ### **Final Answer for 6:** \[ \boxed{v \approx 3.8 \times 10^5 \ \text{m/s}} \] --- ## **7. Work Done by a Force** **Given:** - Force: \( \vec{F} = (2\, \text{N})\hat{i} + (3\, \text{N})\hat{j} \) - Initial position: \( \vec{r}_i = (2\, \text{m})\hat{i} + (3\, \text{m})\hat{j} \) - Final position: \( \vec{r}_f = (3\, \text{m})\hat{i} + (1\, \text{m})\hat{j} \) --- ### **Step 1: Find the Displacement Vector \( \Delta \vec{r} \)** Calculate the displacement: \[ \Delta \vec{r} = \vec{r}_f - \vec{r}_i = [(3-2)\hat{i} + (1-3)\hat{j}] = (1\hat{i} - 2\hat{j})\, \text{m} \] **Explanation:** The displacement vector is found by subtracting the initial position vector from the final position vector, resulting in a vector pointing from the initial to the final position. --- ### **Step 2: Calculate the Work Done \( W \)** Using the dot product: \[ W = \vec{F} \cdot \Delta \vec{r} = (2\hat{i} + 3\hat{j}) \cdot (1\hat{i} - 2\hat{j}) \] Calculating the dot product: \[ W = (2 \times 1) + (3 \times -2) = 2 - 6 = -4\, \text{J} \] **Explanation:** Work is computed as the dot product of the force and displacement vectors, which involves multiplying the corresponding components and summing the results. --- ### **Final Answer for 7:** \[ \boxed{W = -4\, \text{J}} \]