Answer:
0.26
Explanation:
The radius of the table is 2.5 m.
Tangential velocity,
[tex]v=u+at\\u=0\\t=10s\\a=0.25m/s^2\\v=0+0.25\times 10 = 2.5 m/s[/tex]
Net force on the trunk in the plane of the table:
[tex]F=m\sqrt{a^2+(\frac{v^2}{r})^2}\\F= m\sqrt{0.25^2+(\frac{2.5^2}{2.5})^2}=2.5m[/tex]
The trunk slides because the frictional force equals the net force acting on the table:
[tex]\mu m g = 2.5 m\\\mu =\frac{2.5}{g} = 0.26[/tex]
