Calculating Potential and Kinetic Energy for Objects in Motion

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SUMMARY

This discussion focuses on calculating potential and kinetic energy for objects in motion, specifically using examples involving an elevator, colliding cars, and a falling ball. The calculations demonstrate the application of the formulas for kinetic energy (KE = ½ mv²) and potential energy (PE = mgh) in various scenarios. Key results include the height the elevator carried the man (149.64 m), the combined speed of the colliding cars (101.875 km/h), and the energy calculations for the falling ball at different heights. The importance of unit consistency is emphasized, particularly in kinetic energy calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of potential energy and kinetic energy
  • Basic knowledge of unit conversions (e.g., kg to g, km/h to m/s)
  • Proficiency in algebra for solving equations
NEXT STEPS
  • Study the principles of conservation of momentum in collisions
  • Explore advanced applications of energy conservation in mechanical systems
  • Learn about the effects of air resistance on falling objects
  • Investigate real-world applications of energy calculations in engineering
USEFUL FOR

Students in physics, engineers, and anyone interested in understanding the dynamics of motion and energy transformations in physical systems.

phiber
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1. A 2150-kg elevator carries a 90-kg man from the ground up to the top floor of a building in 45 seconds. In doing so, the elevator develops 73000 W of power. How high was the man carried?

(73000 W) (45 s) = 3285 kJ
(9.8 m/s ^2) (2240 kg) = 21952 N

3285 kJ / 21952 N = 149.64 m



2. A 600-kg car traveling at 163 km/hr collides head-on with a 1000-kg parked car. If the cars stick together after impact, how fast are they then traveling? What is their kinetic energy?

(1000 kg) (163 km/h) = (600 kg + 1000 kg) (101.875 km/h)
½ (1600 kg) (101.875) = 8302812 J





3. A 500-g ball rest on top of a tower, 15 meters tall. Calculate its potential and kinetic energies a) when the ball rests on top of the tower, b) after the ball has fallen 10 meters toward the ground, c) 1 meter above the ground.

a. KE = 0
PE = (.5 kg) (9.8 m/s^2) (15 m)
PE = 73.5 J

b. KE = ½ (.5 kg) (196 m/s)
KE = 49 J
PE = (.5 kg) (9.8 m/s^2) (5 m)
PE = 24.5 J

c. KE = ½ (.5 kg) (274.4 m/s)
KE = 68.6 J
PE = (.5 kg) (9.8 m/s^2) (1 m)
PE = 4.9 J
 
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The numbers appear to be ok.

You need to be more careful with your units

KE = ½ (.5 kg) (196 m/s)
Should read

KE = ½ (.5 kg) (196 (m/s)2)
 
seems ok to me...

marlon
 

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