In the lab you measured the velocity of the cart mass m =0.507kg (and negligible

Question

In the lab you measured the velocity of the cart mass m=0.507kg (and negligible error) to be v= 4.15m/s and the standard error of the velocity v= 0.01m/s.

What is the kinetic energy of the cart?


What is the standard error of the kinetic energy of the cart?


Which of the following statements are true regarding your answers?

A. The relative error of kinetic energy equals the relative error of velocity

B. The relative error of kinetic energy is twice the relative error of velocity     

C. The error of kinetic energy is twice the error of velocity

D. The error of kinetic energy equals the error of velocity

Explanation / Answer

a) KE = 1/2mV^2

KE = 1/2*0.507*(4.15)^2 = 4.36 J

b) For error

KE = 1/2mV^2

ln(KE) = ln(m/2) + 2ln(V)   

differentiating this eqution

dKE/KE = dm/m + 2dV/V ( dV , dm are errors in velocity and mass)

dKE = KE*( 0+ 2dV/V) = 4.36*(2*0.01/4.15) = 0.021

c)

relative error is dKE/KE , error is dKE

option B is correct

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