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

Answer:

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

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,

$K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

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

Since, $\omega=\dfrac{v}{r}$

$K_i=mv^2$

$K_i=96\times (0.24)^2=5.529\ J$

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.