speed of the participants is given as
[tex]v = 15 km/h [/tex]
[tex]v = 15 \times \frac{1000 m}{3600s} = 4.2 m/s[/tex]
now the kinetic energy of the participant is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}68\times 4.2^2[/tex]
[tex]KE = 590.3 J[/tex]
As we know that
1 calorie = 4.2 J
[tex]KE = 590.3\times \frac{1 cal}{4.2J} = 141.1 Cal[/tex]
so here 230 Calorie is provided by 1 bar
bar required is
[tex]N = \frac{141.1}{230} = 0.61 bars[/tex]
1 energy bar would be required to fuel such a run for a 68 kg athlete
[tex]\texttt{ }[/tex]
Let's recall Kinetic Energy Formula as follows:
[tex]\large {\boxed{Ek = \frac{1}{2}mv^2} }[/tex]
Ek = Kinetic Energy ( Joule )
m = mass of the object ( kg )
v = speed of the object ( m/s )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
speed of participant = v = 15 km/h = 4¹/₆ m/s
mass of participant = m = 68 kg
Asked:
kinetic energy = Ek = ?
Solution:
Firstly , we will calculate total kinetic energy of the participant:
[tex]Ek = \frac{1}{2}m(v')^2[/tex]
[tex]Ek = \frac{1}{2} \times 68 \times (4\frac{1}{6})^2[/tex]
[tex]Ek = 590 \frac{5}{18} \texttt{ J}[/tex]
[tex]Ek \approx 141 \texttt{ Calories}[/tex]
[tex]\texttt{ }[/tex]
Next , we could find number of 230 Calorie Energy Bars as follows:
[tex]N = Ek / 230[/tex]
[tex]N = 141 / 230[/tex]
[tex]N \approx 0.61[/tex]
1 energy bar would be required to fuel such a run for a 68 kg athlete
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Dynamics
