Answer:
0.05
Explanation:
First, we need to know that the normal force is equal to the weight, so the force of friction is equal to
Ff = μFn
Ff = μmg
Where μ is the coefficient of friction, m is the mass and g is the gravity.
Then, by the work-energy theorem, we get
[tex]\begin{gathered} W=K_f-K_i \\ F_fd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{gathered}[/tex]Where d is the distance traveled, Ff is the force of friction, m is the mass, vf is the final velocity and vi is the initial velocity. The final velocity is 0, so the expression can be simplified as
[tex]\begin{gathered} F_fd=-\frac{1}{2}mv_i^2 \\ \\ \mu mgd=\frac{1}{2}mv_i^2 \end{gathered}[/tex]Solving for μ, we get:
[tex]\begin{gathered} \mu gd=\frac{1}{2}v_i^2 \\ \\ \mu=\frac{v_i^2}{2gd} \end{gathered}[/tex]Replacing vi = 4 m/s, g = 9.8 m/s², and d = 15 m, we get
[tex]\mu=\frac{(4\text{ m/s\rparen}^2}{2(9.8\text{ m/s}^2)(15\text{ m\rparen}}=0.05[/tex]Therefore, the coefficient of sliding friction is 0.05
