A roller coaster at the top of a hill has a potential energy of 5000 J. If it descends to a height of 10 m, what is its speed at the bottom? (g = 9.8 m/s²)
Practice Questions
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Q1
A roller coaster at the top of a hill has a potential energy of 5000 J. If it descends to a height of 10 m, what is its speed at the bottom? (g = 9.8 m/s²)
10 m/s
20 m/s
30 m/s
40 m/s
Using conservation of energy, initial PE + KE = final PE + KE. 5000 J = mgh + 0.5mv². Solving gives v = √(2(5000 - mgh)/m) = 30 m/s.
Questions & Step-by-step Solutions
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Q
Q: A roller coaster at the top of a hill has a potential energy of 5000 J. If it descends to a height of 10 m, what is its speed at the bottom? (g = 9.8 m/s²)
Solution: Using conservation of energy, initial PE + KE = final PE + KE. 5000 J = mgh + 0.5mv². Solving gives v = √(2(5000 - mgh)/m) = 30 m/s.
Steps: 12
Step 1: Understand that potential energy (PE) at the top of the hill is given as 5000 J.
Step 2: Recognize that when the roller coaster descends, its potential energy decreases and converts into kinetic energy (KE).
Step 3: Use the formula for potential energy: PE = mgh, where m is mass, g is acceleration due to gravity (9.8 m/s²), and h is height (10 m).
Step 4: Calculate the potential energy at the height of 10 m: PE = m * 9.8 * 10.
Step 5: Set up the conservation of energy equation: Initial PE (5000 J) = PE at 10 m + KE at the bottom (0.5mv²).
Step 6: Substitute the potential energy at 10 m into the equation: 5000 J = m * 9.8 * 10 + 0.5mv².
Step 7: Rearrange the equation to solve for v: 5000 J - (m * 9.8 * 10) = 0.5mv².
Step 8: Factor out m from the left side: 5000 J - 98m = 0.5mv².
Step 9: Divide both sides by m (assuming m is not zero): (5000 J/m - 98) = 0.5v².
Step 10: Multiply both sides by 2 to eliminate the 0.5: 2(5000 J/m - 98) = v².
Step 11: Take the square root of both sides to find v: v = √(2(5000 J/m - 98)).
Step 12: Since we don't know m, we can simplify the equation to find v in terms of m, but we can also find a numerical answer if we assume a mass.
