1 Answer

Fiw · Answer 1 · 2023-09-23T22:39:52+0000

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

$\mu=\text{ 0.6 }$

(a) The change in internal energy is given by the formula

$\begin{gathered} \Delta U=\text{decrease in kinetic energy} \\ =(1)/(2)m(v)^2-(1)/(2)m(u)^2 \end{gathered}$

Substituting the values, the internal energy will be

$\begin{gathered} \Delta U=(1)/(2)*82.5*0-(1)/(2)*82.5*(3.68)^2 \\ =-\text{ 558.62 J} \end{gathered}$

(b) The distance is given by the formula

$d=(u^2)/(2\mu g)$

Here, the acceleration due to gravity, g= 9.8 m/s^2.

Substituting the values, the distance will be

$\begin{gathered} d=((3.68)^2)/(2*0.6*9.8) \\ =1.51\text{ m} \end{gathered}$

Thus the distance is 1.51 m.

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