Problem concerned to kinematics

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SUMMARY

The discussion focuses on calculating the maximum error in the estimate of kinetic energy, given percentage errors in mass and speed measurements of 2% and 3%, respectively. The derived maximum error in kinetic energy is 8%, based on the equation K = mv²/2. The relationship between the changes in kinetic energy and the measurements is established through the formula dK/K = dm/m + 2dv/v, which is derived from differentiating the logarithmic form of the kinetic energy equation.

PREREQUISITES
  • Understanding of basic kinematics and the kinetic energy formula K = mv²/2
  • Familiarity with calculus concepts, specifically differentiation
  • Knowledge of error analysis in measurements
  • Basic understanding of logarithmic functions and their properties
NEXT STEPS
  • Study the principles of error propagation in physics
  • Learn about differentiation techniques in calculus
  • Explore the application of logarithmic differentiation in various equations
  • Investigate the implications of measurement errors on physical calculations
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This discussion is beneficial for physics students, educators, and professionals involved in experimental physics, particularly those focusing on kinematics and error analysis in measurements.

Mareena
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The percentage errors in the measurement of mass and speed are 2 % and 3% respectively.how much wil be the maximum error in the estimate of kinetic energy obtained by measuring mass and speed?
its ans is 8 % ..
according to this equation K = mv^2/ 2 , so dK / K = dm/m + 2dv / v
( d means delta ) .. but the problem is , where dat second equation has been derived from? please help
 
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Hi Mareena! :smile:

(please don't use txt speak on this forum :redface:)
Mareena said:
… K = mv^2/ 2 , so dK / K = dm/m + 2dv / v
( d means delta ) .. but the problem is , where dat second equation has been derived from?

it comes from differentiating logK = logm + 2logv + logconstant :wink:
 

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