Question

A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it

Answer 1

The work done by the hoop is equal to 5.529 Joules.

Step-by-step explanation:

Given that,

Mass of the hoop, m = 96 kg

The speed of the center of mass, v = 0.24 m/s

To find,

The work done by the hoop.

Solution,

The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,

`K_i=(1)/(2)mv^2+(1)/(2)I\omega^2`

I is the moment of inertia,
`I=mr^2`

Since,
`\omega=(v)/(r)`

`K_i=mv^2`

`K_i=96* (0.24)^2=5.529\ J`

Finally it stops, so the final energy of the hoop will be,
`K_f=0`

The work done by the hoop is equal to the change in kinetic energy as :

`W=K_f-K_i`

W = -5.529 Joules

So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.