2. A loaded barge has a mass of 1500000 kg and is traveling at 3m/s If a tugboat applies an opposing force of 12000 N for 10 s, what is its final velocity? How long will it take to stop the barge?How much force would a tugboat need to apply to stop it in one minute?

To solve this problem, we can use the concepts of momentum and Newton's second law.<br /><br />1. **Final velocity after 10 seconds:**<br /><br /> The initial momentum of the barge is given by:<br /> \[<br /> p_{\text{initial}} = m \cdot v_{\text{initial}} = 1500000 \, \text{kg} \times 3 \, \text{m/s} = 4500000 \, \text{kg m/s}<br /> \]<br /><br /> The force applied by the tugboat is 12000 N in the opposite direction, so the change in momentum over 10 seconds is:<br /> \[<br /> \Delta p = F \cdot t = 12000 \, \text{N} \times 10 \, \text{s} = 120000 \, \text{kg m/s}<br /> \]<br /><br /> The final momentum is:<br /> \[<br /> p_{\text{final}} = p_{\text{initial}} - \Delta p = 4500000 \, \text{kg m/s} - 120000 \, \text{kg m/s} = 4380000 \, \text{kg m/s}<br /> \]<br /><br /> The final velocity \( v_{\text{final}} \) is:<br /> \[<br /> v_{\text{final}} = \frac{p_{\text{final}}}{m} = \frac{4380000 \, \text{kg m/s}}{1500000 \, \text{kg}} = 2.92 \, \text{m/s}<br /> \]<br /><br />2. **Time to stop the barge:**<br /><br /> To find the time it takes to stop the barge, we need to calculate how long it takes for the velocity to reach zero with a constant opposing force of 12000 N.<br /><br /> Using the formula \( F = m \cdot a \), where \( a \) is the acceleration:<br /> \[<br /> a = \frac{F}{m} = \frac{12000 \, \text{N}}{1500000 \, \text{kg}} = 0.008 \, \text{m/s}^2<br /> \]<br /><br /> The time \( t \) to stop the barge from an initial velocity of 3 m/s is:<br /> \[<br /> v_{\text{final}} = v_{\text{initial}} + a \cdot t<br /> \]<br /> \[<br /> 0 = 3 \, \text{m/s} - 0.008 \, \text{m/s}^2 \cdot t<br /> \]<br /> \[<br /> t = \frac{3 \, \text{m/s}}{0.008 \, \text{m/s}^2} = 375 \, \text{s}<br /> \]<br /><br />3. **Force needed to stop the barge in one minute (60 seconds):**<br /><br /> If the barge needs to be stopped in 60 seconds, we use the same formula for acceleration:<br /> \[<br /> v_{\text{final}} = v_{\text{initial}} + a \cdot t<br /> \]<br /> \[<br /> 0 = 3 \, \text{m/s} + a \cdot 60 \, \text{s}<br /> \]<br /> \[<br /> a = -\frac{3 \, \text{m/s}}{60 \, \text{s}} = -0.05 \, \text{m/s}^2<br /> \]<br /><br /> The required force \( F \) is:<br /> \[<br /> F = m \cdot a = 1500000 \, \text{kg} \times (-0.05 \, \text{m/s}^2) = -75000 \, \text{N}<br /> \]<br /><br /> Therefore, a force of 75000 N in the opposite direction is needed to stop the barge in one minute.