A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?
Practice Questions
1 question
Q1
A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?
5 m/s
10 m/s
15 m/s
20 m/s
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.
Questions & Step-by-step Solutions
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Q
Q: A 1 kg mass is dropped from a height of 10 m. What is the speed just before it hits the ground?
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.
Steps: 12
Step 1: Identify the mass of the object. In this case, the mass is 1 kg.
Step 2: Identify the height from which the object is dropped. Here, the height is 10 m.
Step 3: Understand that when the object is dropped, it has potential energy at the height and will convert that energy into kinetic energy just before it hits the ground.
Step 4: Write the formula for potential energy (PE) at height: PE = mgh, where m is mass, g is acceleration due to gravity (approximately 9.8 m/s²), and h is height.
Step 5: Write the formula for kinetic energy (KE) just before hitting the ground: KE = 0.5mv², where v is the speed just before impact.
Step 6: Set the potential energy equal to the kinetic energy: mgh = 0.5mv².
Step 7: Since mass (m) is on both sides of the equation, you can cancel it out: gh = 0.5v².
Step 8: Rearrange the equation to solve for v²: v² = 2gh.
Step 9: Substitute the values for g (9.8 m/s²) and h (10 m) into the equation: v² = 2 * 9.8 * 10.
Step 10: Calculate the value: v² = 196.
Step 11: Take the square root of both sides to find v: v = sqrt(196).
