What Is the Kinetic Energy of a 950-lb Motorcycle Moving at 50 mph?

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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 a value of 1.1 × 105 joules. When the speed decreases to 25 mph, the ratio of the kinetic energies K1 (at 50 mph) to K2 (at 25 mph) is determined to be 0.25. This indicates that K2 is one-fourth of K1, demonstrating the quadratic relationship between speed and kinetic 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 how speed affects kinetic energy
NEXT STEPS
  • Research unit conversions for mass and speed in physics
  • Learn about the implications of kinetic energy in real-world scenarios
  • Explore the relationship between speed and kinetic energy in different contexts
  • Study the effects of mass on kinetic energy calculations
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of kinetic energy and motion dynamics.

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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?


Homework Equations



1. KE=1/2mv2
The answer I got for this problem was 1.1×10 5 J

The Attempt at a Solution



2.k1/k2= 0.25
 
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Yeah, those seem about right...
 

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