A 8.0-kg mass moves, starting from rest at \( x=0 \). An initial, \( F=(3.0 \mathrm{~N}) i \), force is applied as the mass moves from \( x=0.0 \) to \( x=8.0 \mathrm{~m} \). A second, \( F=(-5.0 \mathrm{~N}) \mathrm{i} \), force is applied later as the mass moves from \( x=10 \mathrm{~m} \) to \( x=12 \mathrm{~m} \). What is the mass's speed as it passes \( x=12 \mathrm{~m} \) ?

To find the mass's speed as it passes \( x = 12 \, \text{m} \), we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. 1. **Calculate the work done by the first force** (\( F_1 = 3.0 \, \text{N} \)) as the mass moves from \( x = 0 \, \text{m} \) to \( x = 8 \, \text{m} \): \[ W_1 = F_1 \cdot d_1 = 3.0 \, \text{N} \cdot (8.0 \, \text{m} - 0.0 \, \text{m}) = 3.0 \, \text{N} \cdot 8.0 \, \text{m} = 24.0 \, \text{J} \] 2. **Calculate the work done by the second force** (\( F_2 = -5.0 \, \text{N} \)) as the mass moves from \( x = 10 \, \text{m} \) to \( x = 12 \, \text{m} \): \[ W_2 = F_2 \cdot d_2 = -5.0 \, \text{N} \cdot (12.0 \, \text{m} - 10.0 \, \text{m}) = -5.0 \, \text{N} \cdot 2.0 \, \text{m} = -10.0 \, \text{J} \] 3. **Calculate the total work done on the mass**: \[ W_{\text{total}} = W_1 + W_2 = 24.0 \, \text{J} - 10.0 \, \text{J} = 14.0 \, \text{J} \] 4. **Use the work-energy principle to find the final speed**. The initial kinetic energy (\( KE_i \)) is zero since the mass starts from rest. The final kinetic energy (\( KE_f \)) can be expressed as: \[ KE_f = \frac{1}{2} m v^2 \] Setting the total work equal to the change in kinetic energy: \[ W_{\text{total}} = KE_f - KE_i \] \[ 14.0 \, \text{J} = \frac{1}{2} (8.0 \, \text{kg}) v^2 - 0 \] 5. **Solve for \( v \)**: \[ 14.0 \, \text{J} = \frac{1}{2} (8.0 \, \text{kg}) v^2 \] \[ 14.0 = 4.0 v^2 \] \[ v^2 = \frac{14.0}{4.0} = 3.5 \] \[ v = \sqrt{3.5} \approx 1.87 \, \text{m/s} \] Thus, the mass's speed as it passes \( x = 12 \, \text{m} \) is approximately \( 1.87 \, \text{m/s} \).