A 3 kg object is dropped from a height of 12 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

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Q: A 3 kg object is dropped from a height of 12 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)

Solution: Using conservation of energy, PE = KE: mgh = 1/2 mv²; v = sqrt(2gh) = sqrt(2 * 10 m/s² * 12 m) = 15.49 m/s

Steps: 12

Step 1: Identify the mass of the object (m = 3 kg) and the height from which it is dropped (h = 12 m).
Step 2: Use the acceleration due to gravity (g = 10 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 formula for potential energy: PE = mgh.
Step 5: Write the formula for kinetic energy: KE = 1/2 mv².
Step 6: Set the potential energy equal to the kinetic energy: mgh = 1/2 mv².
Step 7: Notice that the mass (m) cancels out from both sides of the equation.
Step 8: Rearrange the equation to solve for v (speed): v² = 2gh.
Step 9: Substitute the values of g and h into the equation: v² = 2 * 10 m/s² * 12 m.
Step 10: Calculate the right side: v² = 240 m²/s².
Step 11: Take the square root of both sides to find v: v = sqrt(240 m²/s²).
Step 12: Calculate the square root: v ≈ 15.49 m/s.