A ball rolls down a ramp of height h. If it starts from rest, what is its final speed at the bottom?

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Q: A ball rolls down a ramp of height h. If it starts from rest, what is its final speed at the bottom?

Solution: Using conservation of energy, potential energy mgh converts to kinetic energy (1/2 mv^2). Thus, v = √(2gh).

Steps: 9

Step 1: Understand that the ball starts at a height 'h' and has potential energy due to its height.
Step 2: Remember the formula for potential energy (PE) which is PE = mgh, where 'm' is mass, 'g' is acceleration due to gravity, and 'h' is height.
Step 3: When the ball rolls down, this potential energy converts into kinetic energy (KE).
Step 4: The formula for kinetic energy is KE = (1/2)mv^2, where 'v' is the final speed of the ball.
Step 5: Set the potential energy equal to the kinetic energy: mgh = (1/2)mv^2.
Step 6: Notice that 'm' (mass of the ball) can be canceled from both sides of the equation since it is present in both terms.
Step 7: You are left with the equation: gh = (1/2)v^2.
Step 8: To solve for 'v', multiply both sides by 2: 2gh = v^2.
Step 9: Take the square root of both sides to find 'v': v = √(2gh).