Problem concerned to kinematics

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The discussion centers on calculating the maximum error in kinetic energy estimation due to percentage errors in mass and speed measurements, which are 2% and 3%, respectively, resulting in an 8% maximum error. The formula for kinetic energy, K = mv^2/2, is referenced, and the relationship between changes in kinetic energy and the errors in mass and speed is expressed as dK/K = dm/m + 2dv/v. A participant seeks clarification on the derivation of this equation. It is explained that the equation is derived from differentiating the logarithmic form of the kinetic energy equation. Understanding this derivation is crucial for accurately assessing measurement errors in physics.
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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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