A 2 kg object is dropped from a height of 10 m. What is its speed just before it

A 2 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A 2 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

Step 1: Identify the mass of the object (m = 2 kg) and the height from which it is dropped (h = 10 m).
Step 2: Use the acceleration due to gravity (g = 9.8 m/s²).
Step 3: Understand that when the object is dropped, its potential energy (PE) at the height will convert to kinetic energy (KE) just before it hits the ground.
Step 4: Write the equation for potential energy: PE = mgh.
Step 5: Write the equation for kinetic energy: KE = 0.5mv².
Step 6: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 7: Notice that the mass (m) can be canceled from both sides of the equation since it is the same on both sides.
Step 8: Simplify the equation to: gh = 0.5v².
Step 9: Rearrange the equation to solve for v²: v² = 2gh.
Step 10: Substitute the values of g and h into the equation: v² = 2 × 9.8 m/s² × 10 m.
Step 11: Calculate the right side: v² = 196 m²/s².
Step 12: Take the square root of both sides to find v: v = √(196 m²/s²).
Step 13: Calculate the square root: v = 14 m/s.

Energy Conservation – The principle that the total mechanical energy (potential + kinetic) in a closed system remains constant if only conservative forces are acting.
Kinetic and Potential Energy – Understanding the relationship between potential energy (PE = mgh) and kinetic energy (KE = 0.5mv²) in the context of free fall.
Free Fall Motion – The motion of an object under the influence of gravity alone, without any air resistance.