Question

What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest?

Answer 1

Answer: The final speed of the hoop will be
`7√(2)m/s`

Step-by-step 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:

`mgh=(1)/(2)mv^2`

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:

`m* 9.8* 5=(1)/(2)* m* v^2\\\\v^2=9.8* 5* 2\\\\v=√(98)=7√(2)m/s`

Hence, the final speed of the hoop will be
`7√(2)m/s`

Answer 2

In every natural phenomena existing, there are three law of conservations that must always be follows: the Law of Conservation of Mass, Energy and Momentum. Let's focus particularly in the conservation of energy. Energy is neither created nor destroyed, it just transferred from one form to another. With that being said, the energy from the hoop from start to finish does not change. When ti was on top of the hill, it possesses potential energy. As it moves down the hill, potential energy is transformed to kinetic energy. So, the conservation is presented in this equation:

Potential energy = Kinetic energy
mgh = 1/2*mv²
where
m is the mass of the hoop
g is the acceleration due to gravity equal to 9.81 m/s²
h is the height
v is the final velocity

Simplifying the equation, we cancel out m at both sides:
gh = 1/2*v²
Substituting the values:
(9.81)(5) = 1/2*v²
v² = 98.1
v = √98.1
v = 9.9 m/s