Answer:
Explanation:
Given
mass of object [tex]m=2 kg[/tex]
inclination [tex]\theta =50^{\circ}[/tex]
[tex]\mu _k=0.3[/tex]
[tex]\mu _s=0.4[/tex]
initial velocity [tex]u=3 m/s[/tex]
acceleration of block during upward motion
[tex]a=g\sin \theta -\mu _kg\cos \theta [/tex]
[tex]a=g(\sin 50-0.3\cos 50)[/tex]
[tex]a=5.617 m/s^2[/tex]
using relation
[tex]v^2-u^2=2a\cdot s[/tex]
where [tex]s=distance\ moved [/tex]
[tex]v=final\ velocity[/tex]
v=0 because block stopped after moving distance s
[tex]0-(3)^2=2\cdot (-5.617)\cdot s[/tex]
[tex]s=\frac{4.5}{5.617}[/tex]
[tex]s=0.801[/tex]
If block stopped after s m then force acting on block is
[tex]F=mg\sin \theta =[/tex]friction force [tex]f_r=\mu mg\cos \theta [/tex]
[tex]F>f_r[/tex] therefore block will slide back down to the bottom
