A 2 kg object is dropped from a height of 5 m. What is its speed just before it

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

A 2 kg object is dropped 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 object (m = 2 kg) and the height from which it is dropped (h = 5 m).
Step 2: Use the acceleration due to gravity (g = 9.8 m/s²).
Step 3: Write the formula for potential energy (PE) at the height: PE = mgh.
Step 4: Write the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5mv².
Step 5: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 6: Notice that the mass (m) cancels out from both sides of the equation.
Step 7: Rearrange the equation to solve for v: gh = 0.5v².
Step 8: Multiply both sides by 2 to eliminate the 0.5: 2gh = v².
Step 9: Take the square root of both sides to find v: v = √(2gh).
Step 10: Substitute the values for g and h into the equation: v = √(2 × 9.8 m/s² × 5 m).
Step 11: Calculate the value: v = √(98) which is approximately 10 m/s.

Conservation of Energy – This principle states that the total energy in a closed system remains constant. In this case, the potential energy of the object at height is converted into kinetic energy just before it hits the ground.
Kinetic and Potential Energy – Understanding the relationship between kinetic energy (energy of motion) and potential energy (stored energy due to position) is crucial for solving problems involving falling objects.
Gravitational Acceleration – The acceleration due to gravity (g = 9.8 m/s²) is a key factor in calculating the speed of the object just before impact.