What is the kinetic energy in joules of a 950-lb motorcycle moving

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SUMMARY

The kinetic energy of a 950-lb motorcycle moving at 50 mph is calculated using the formula KE = 1/2 mv², resulting in approximately 1.1 × 10⁵ joules. When the speed decreases to 25 mph, the kinetic energy ratio K1 to K2 is determined to be 0.25, indicating that K2 is one-fourth of K1. This analysis highlights the significant impact of velocity on kinetic energy, demonstrating the quadratic relationship between speed and energy.

PREREQUISITES
  • Understanding of kinetic energy formula (KE = 1/2 mv²)
  • Basic knowledge of unit conversions (pounds to kilograms, mph to m/s)
  • Familiarity with ratios and their calculations
  • Concept of energy in physics
NEXT STEPS
  • Research unit conversion techniques for mass and speed
  • Explore the implications of kinetic energy in real-world scenarios
  • Learn about energy conservation principles in physics
  • Investigate the effects of speed on kinetic energy in different contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of kinetic energy calculations.

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1. Homework Statement
1. What is the kinetic energy in joules of a 950-lb motorcycle moving at 50 mph ?

2. If the speed of the motorcycle changes from 50 mph (with kintic energy K1) to 25 mph (with kinetic energy K2), what is the ratio of K1 to K2?


2. Homework Equations

KE=1/2mv2


3. The Attempt at a Solution
1. 1.1×10 5 J
2. k1/k2= 0.25
 
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You already posted this in the homework section, this area isn't for homework problems, and this problem has nothing to do with chemistry.

You hit the trifecta.
 

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