If a 2 kg object is dropped from a height of 5 m, what is its speed just before

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Question: If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)

Options:

Correct Answer: 10 m/s

Solution:

Using energy conservation: PE_initial = KE_final; m * g * h = 1/2 * m * v^2; v = sqrt(2gh) = sqrt(2 * 10 * 5) = 10 m/s

If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)

If a 2 kg object is dropped from a height of 5 m, what is its speed just before it hits the ground? (g = 10 m/s²)

Step 1: Identify the mass of the object. Here, the mass (m) is 2 kg.
Step 2: Identify the height from which the object is dropped. Here, the height (h) is 5 m.
Step 3: Identify the acceleration due to gravity. Here, g is given as 10 m/s².
Step 4: Use the formula for potential energy (PE) at the height: PE_initial = m * g * h.
Step 5: Calculate the potential energy: PE_initial = 2 kg * 10 m/s² * 5 m = 100 Joules.
Step 6: Use the formula for kinetic energy (KE) just before hitting the ground: KE_final = 1/2 * m * v².
Step 7: Set the potential energy equal to the kinetic energy: PE_initial = KE_final.
Step 8: Substitute the values: 100 Joules = 1/2 * 2 kg * v².
Step 9: Simplify the equation: 100 = 1 kg * v².
Step 10: Solve for v²: v² = 100.
Step 11: Take the square root to find v: v = sqrt(100) = 10 m/s.

Energy Conservation – The principle that the total energy in a closed system remains constant, allowing potential energy to convert into kinetic energy.
Kinematics – The study of motion, particularly how speed and acceleration relate to distance and time.
Gravitational Acceleration – The acceleration due to gravity, which is approximately 10 m/s² near the Earth's surface.