5. Gary (60kg) gives Joanne (40kg) a 1900 N push with his really long arms. How far apart would they be after 0.5 seconds if there is a coefficient of friction of 01?

Here's how to solve this problem:<br /><br />1. **Calculate the net force on each person:**<br /><br />* **Gary:** Gary pushes Joanne with 1900 N. Since every action has an equal and opposite reaction (Newton's Third Law), Joanne exerts a 1900 N force back on Gary. The force of friction on Gary is calculated as: Friction = (coefficient of friction) * (Normal Force). Gary's normal force is equal to his weight (mass * gravity) = 60 kg * 9.8 m/s² = 588 N. Therefore, the friction on Gary is 0.1 * 588 N = 58.8 N. The net force on Gary is 1900 N (push) - 58.8 N (friction) = 1841.2 N.<br /><br />* **Joanne:** Joanne experiences the 1900 N push from Gary. The force of friction on Joanne is calculated similarly: Friction = 0.1 * (40 kg * 9.8 m/s²) = 39.2 N. The net force on Joanne is 1900 N (push) - 39.2 N (friction) = 1860.8 N.<br /><br />2. **Calculate the acceleration of each person:**<br /><br />* **Gary:** Acceleration = Net Force / Mass = 1841.2 N / 60 kg = 30.687 m/s²<br /><br />* **Joanne:** Acceleration = Net Force / Mass = 1860.8 N / 40 kg = 46.52 m/s²<br /><br />3. **Calculate the distance each person travels in 0.5 seconds:**<br /><br />We'll use the equation of motion: distance = initial velocity * time + 0.5 * acceleration * time²<br /><br />Since they start from rest, their initial velocity is 0.<br /><br />* **Gary:** distance = 0.5 * 30.687 m/s² * (0.5 s)² = 3.836 m<br /><br />* **Joanne:** distance = 0.5 * 46.52 m/s² * (0.5 s)² = 5.815 m<br /><br />4. **Calculate the distance between Gary and Joanne:**<br /><br />Since they are pushed in opposite directions, the total distance between them is the sum of the distances they each travel: 3.836 m + 5.815 m = 9.651 m<br /><br />**Therefore, Gary and Joanne will be approximately 9.65 meters apart after 0.5 seconds.**<br />