Good morning.
We calculate the acceleration with the Torricelli equation:
[tex]\star\ \boxed{\mathsf{V^2 = V_0^2 + 2a\Delta S}}[/tex]
We see that:
[tex]\begin{cases}\mathsf{V_0 = 0}\\\mathsf{V = 2.3 \ m/s}\\\mathsf{\Delta S = 8.9 \ m}\end{cases}[/tex]
Now:
[tex]\mathsf{(2.3)^2 = 0^2 + 2a\cdot 8.9}\\ \\ \mathsf{5.29 = 17.8a}\\ \\ \bold{\mathsf{a = 0.3 \ m/s^2}}[/tex]
Now we can calculate the resultant force that makes that acceleration of 0.3 m/s² with the 2nd Law of Newton:
[tex]\mathsf{F_r = m\cdot a}\\ \\ \mathsf{F_r = 20\cdot 0.3}\\ \\ \mathsf{F_r = 6 \ N}[/tex]
We have a force of 26 N → and a friction force F ←. Adding those vectors, he have a force 6 N →. Therefore:
26 - F = 6
F = 20 N
We have a friction force of 20 N. We calculate the kinect coefficient with the formula:
[tex]\star \ \boxed{\mathsf{F = \mu_k N}}[/tex]
Since we are in a horizontal plane, we hava that N = P = mg = 200 N
Therefore:
[tex]\mathsf{20 = \mu_k 200}\\ \\ \boxed{\boxed{\mathsf{\mu_k = 0.1}}}[/tex]
