A block of mass 2 kg is released from a height of 5 m. What is its speed just be

A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)

Step 1: Identify the mass of the block, which is 2 kg.
Step 2: Identify the height from which the block is released, which is 5 m.
Step 3: Identify the acceleration due to gravity, which is 9.8 m/s².
Step 4: Use the formula for potential energy (PE) at height: PE = mgh, where m is mass, g is gravity, and h is height.
Step 5: Calculate the potential energy: PE = 2 kg * 9.8 m/s² * 5 m.
Step 6: Calculate the potential energy: PE = 98 Joules.
Step 7: Use the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5mv².
Step 8: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 9: Since mass (m) is on both sides, it can be canceled out: gh = 0.5v².
Step 10: Rearrange the equation to solve for v: v² = 2gh.
Step 11: Substitute the values of g and h into the equation: v² = 2 * 9.8 m/s² * 5 m.
Step 12: Calculate v²: v² = 98 m²/s².
Step 13: Take the square root to find v: v = √(98) m/s.
Step 14: Calculate the final speed: v ≈ 9.9 m/s, which can be rounded to 10 m/s.

Conservation of Energy – The principle that energy cannot be created or destroyed, only transformed from one form to another, in this case from potential energy to kinetic energy.
Kinetic Energy – The energy an object possesses due to its motion, calculated as 0.5mv².
Potential Energy – The energy stored in an object due to its height above the ground, calculated as mgh.