Answer: The final speed of the hoop will be [tex]7\sqrt{2}m/s[/tex]
Explanation:
As, the hoop is rolling down the hill which is 5 m high, ts initial potential energy is getting converted to the kinetic energy because the energy is conserved in the system.
Law of conservation of energy states that energy can neither be created nor be destroyed, but it can only be transformed from one form to another form. Here, potential energy is getting transformed from potential energy to kinetic energy.
So, writing the equation for conservation of mass, we get:
[tex]mgh=\frac{1}{2}mv^2[/tex]
where,
m = mass of the hoop
h = height of the hoop = 5 m
g = acceleration due to gravity = 9.8 m/s
v = final velocity of the hoop = ? m/s
Putting values in above equation, we get:
[tex]m\times 9.8\times 5=\frac{1}{2}\times m\times v^2\\\\v^2=9.8\times 5\times 2\\\\v=\sqrt{98}=7\sqrt{2}m/s[/tex]
Hence, the final speed of the hoop will be [tex]7\sqrt{2}m/s[/tex]
