Given that the initial speed, u = 3.68 m/s^2 anf the final speed is v=0.
and the mass of the player, m = 82.5 kg, and the coefficient of friction is
[tex]\mu=\text{ 0.6 }[/tex]
(a) The change in internal energy is given by the formula
[tex]\begin{gathered} \Delta U=\text{decrease in kinetic energy} \\ =\frac{1}{2}m(v)^2-\frac{1}{2}m(u)^2 \end{gathered}[/tex]
Substituting the values, the internal energy will be
[tex]\begin{gathered} \Delta U=\frac{1}{2}\times82.5\times0-\frac{1}{2}\times82.5\times(3.68)^2 \\ =-\text{ 558.62 J} \end{gathered}[/tex]
(b) The distance is given by the formula
[tex]d=\frac{u^2}{2\mu g}[/tex]
Here, the acceleration due to gravity, g= 9.8 m/s^2.
Substituting the values, the distance will be
[tex]\begin{gathered} d=\frac{(3.68)^2}{2\times0.6\times9.8} \\ =1.51\text{ m} \end{gathered}[/tex]
Thus the distance is 1.51 m.
