What is the coefficient of kinetic friction between the block and the surface? Express your answer using two significant figures.
Answer:
0.39
Explanation:
Given information
m1=1.7 Kg
m2=0.011 Kg
v2=670 m/s
d=2.4 m
m2v2=(m1+m2)v hence [tex]v=\frac {m2v2}{m1+m2}[/tex] and also [tex]a=\frac {v^{2}}{2d}[/tex]
The deceleration due to friction is given by
[tex]F=\mu_k N=\mu_k W=\mu_k (m1+m2)g[/tex]
[tex]F=(m1+m2)a=\mu_k (m1+m2)g[/tex] therefore [tex]a=\mu_k g[/tex]
Therefore, [tex]a=\frac {v^{2}}{2d}=\mu_k g[/tex]
[tex]\mu_k=\frac {1}{2dg}(\frac {m2v}{m1+m2})^{2}=\frac {1}{2*2.4*9.81}\times (\frac {0.011*670}{1.7+0.011})^{2}[/tex]
[tex]\mu_k=0.39[/tex]
