A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?
Practice Questions
1 question
Q1
A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?
2 m/s
4 m/s
6 m/s
8 m/s
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.
Questions & Step-by-step Solutions
1 item
Q
Q: A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?
Solution: Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.
Steps: 11
Step 1: Understand that the pendulum swings from a height of 2 meters.
Step 2: Recognize that at the highest point, the pendulum has potential energy and at the lowest point, it has kinetic energy.
Step 3: Use the formula for potential energy (PE) at the top: PE = mgh, where m is mass, g is gravity (9.8 m/s²), and h is height (2 m).
Step 4: At the lowest point, all potential energy converts to kinetic energy (KE), which is given by the formula KE = 0.5mv².
Step 5: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 6: Notice that mass (m) cancels out from both sides of the equation, simplifying it to gh = 0.5v².
Step 7: Rearrange the equation to solve for v²: v² = 2gh.
Step 8: Substitute the values for g (9.8 m/s²) and h (2 m): v² = 2 * 9.8 * 2.
Step 9: Calculate the right side: v² = 39.2.
Step 10: Take the square root of 39.2 to find v: v = √39.2.
Step 11: Calculate the square root to find the speed: v ≈ 6.26 m/s.
