Answer:
The speed when it reaches the lower end of the board is 1.9m/s
Explanation:
We can use here the equation of energy conservation given by,
[tex]mgh = \frac{1}{2} mv^2+\frac{1}{2}I\omega^2[/tex]
Where,
[tex]I= \frac{2}{5}mR^2[/tex]
[tex]\omega = \frac{v}{R}[/tex]
m= mass of sphere
v=velocity of sphere
Replacing the equation in our first formula we have,
[tex]mgh = \frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2[/tex]
[tex]gh = \frac{1}{2}v^2+\frac{1}{5}v^2[/tex]
[tex](9.8)(0.26)=\frac{7}{10}v^2[/tex]
[tex]v=\sqrt{(9.8)(0.26)(10)/7}[/tex]
[tex]v=1.9m/s[/tex]
Therefore the speed when it reaches the lower end of the board is 1.9m/s
