Answer:5.60 m/s
Explanation:
Given
Coefficient of static friction [tex]\mu _s=0.24[/tex]
Coefficient of kinetic friction [tex]\mu _k=0.22[/tex]
mass of crate [tex]m=20\ kg[/tex]
Force applied [tex]F=200\ N[/tex]
maximum static Friction [tex]F_s=\mu _sN[/tex]
[tex]N=mg[/tex]
[tex]F_s=0.24\times 20\times 9.8[/tex]
[tex]F_s=47.04\ N[/tex]
thus applied force is greater than Static friction therefore kinetic friction will come into play
[tex]F_k=\mu _kN[/tex]
[tex]F_k=0.22\times 20\times 9.8=43.12\ N[/tex]
net Force on crate [tex]F-F_k=ma[/tex]
[tex]a=\frac{200-43.12}{20}=7.84\ m/s^2[/tex]
Magnitude of velocity can be obtained by using
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
here initial velocity is zero as crate start from rest
[tex]v^2-0=2\times 7.84\times 2[/tex]
[tex]v=\sqrt{31.37}[/tex]
[tex]v=5.60\ m/s[/tex]
