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A block of mass 2 kg is released from a height of 10 m. What is its speed just b

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Question: A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?

Options:

  1. 0 m/s
  2. 10 m/s
  3. 14 m/s
  4. 20 m/s

Correct Answer: 14 m/s

Solution: 

Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.

A block of mass 2 kg is released from a height of 10 m. What is its speed just b

Practice Questions

Q1
A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?
  1. 0 m/s
  2. 10 m/s
  3. 14 m/s
  4. 20 m/s

Questions & Step-by-Step Solutions

A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?
  • 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 10 m.
  • Step 3: Use the formula for gravitational potential energy (PE) at height: PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s²).
  • Step 4: Calculate the potential energy: PE = 2 kg * 9.8 m/s² * 10 m.
  • Step 5: Simplify the calculation: PE = 196 Joules.
  • Step 6: Understand that just before the block hits the ground, all potential energy converts to kinetic energy (KE).
  • Step 7: Use the formula for kinetic energy: KE = 0.5 * m * v².
  • Step 8: Set the potential energy equal to the kinetic energy: 196 Joules = 0.5 * 2 kg * v².
  • Step 9: Simplify the equation: 196 = 1 * v², so v² = 196.
  • Step 10: Take the square root of both sides to find v: v = sqrt(196).
  • Step 11: Calculate the final speed: v = 14 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 and Potential Energy – Understanding the relationship between potential energy (mgh) and kinetic energy (0.5mv^2) is crucial for solving problems involving falling objects.
  • Gravitational Acceleration – Recognizing that the acceleration due to gravity (g) is approximately 9.8 m/s² near the Earth's surface.
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