A 8.O-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, \( F=(-5.0 \mathrm{~N}) i \), 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 determine the speed of the mass as it passes \( x = 12 \mathrm{~m} \), we can use the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy. 1. **From \( x = 0 \) to \( x = 8 \mathrm{~m} \)**: - The force applied is \( F_1 = 3.0 \mathrm{~N} \). - The work done by this force can be calculated as: \[ W_1 = F_1 \cdot d_1 = 3.0 \, \mathrm{N} \cdot 8 \, \mathrm{m} = 24 \, \mathrm{J} \] 2. **From \( x = 10 \mathrm{~m} \) to \( x = 12 \mathrm{~m} \)**: - The force applied is \( F_2 = -5.0 \mathrm{~N} \) (opposite direction). - The distance moved during this segment is \( d_2 = 2 \, \mathrm{m} \). - The work done by this force is: \[ W_2 = F_2 \cdot d_2 = -5.0 \, \mathrm{N} \cdot 2 \, \mathrm{m} = -10 \, \mathrm{J} \] 3. **Total Work Done**: - The total work done on the mass is: \[ W_{\text{total}} = W_1 + W_2 = 24 \, \mathrm{J} - 10 \, \mathrm{J} = 14 \, \mathrm{J} \] 4. **Change in Kinetic Energy**: - Since the mass starts from rest, the kinetic energy at the beginning (\( KE_i \)) is 0. The final kinetic energy (\( KE_f \)) can be expressed as: \[ KE_f = W_{\text{total}} = \frac{1}{2}mv^2 \] - Plugging in the values: \[ 14 \, \mathrm{J} = \frac{1}{2} \times 8.0 \, \mathrm{kg} \times v^2 \] 5. **Solving for Speed**: \[ 14 = 4.0 \, v^2 \] \[ v^2 = \frac{14}{4.0} = 3.5 \] \[ v = \sqrt{3.5} \approx 1.87 \, \mathrm{m/s} \] Thus, the mass's speed as it passes \( x = 12 \mathrm{~m} \) is approximately **1.87 m/s**.