Answer:
[tex]e=3367.2J[/tex]
%[tex]e=1.43[/tex]%
Explanation:
From the exercise we know two information. The real speed and the experimental measured by the speedometer
[tex]v_{r}=10km/h=2.77m/s[/tex]
Since the speedometer is only accurate to within 0.1km/h the experimental speed is
[tex]v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s[/tex]
Knowing that we can calculate Kinetic energy for the real and experimental speed
[tex]E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J[/tex]
[tex]E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J[/tex]
Now, the potential error in her calculated kinetic energy is:
[tex]e=E_{r}-E_{e}=(234023-230656)J=3367.2J[/tex]
%[tex]e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43[/tex]%
