Question

A 130 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.180 m/s. How much work must be done on the hoop to stop it?

Answer 1

Answer:

The work must be done on the hoop to stop it is 4.21 Joules.

Explanation:

Given that,

Mass of the hoop, m = 130 kg

Initial speed of the hoop, u = 0.18 m/s

Finally it stops, v = 0

We know that the work done by an object is equal to the change in its kinetic energy. Final kinetic energy of the hoop is equal to 0 as the hoop stops.

So,

$W=-(K_t+K_r)$

$K_t\ and\ K_r$ are translational and rotational kinetic energy of the hoop

$W=-(\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2)$

$I=mr^2\ and\ \omega=\dfrac{v}{r}$

$W=-mv^2$

$W=-130\ kg\times (0.18\ m/s)^2$

W = -4.21 Joules

So, the work must be done on the hoop to stop it is 4.21 Joules. Hence, this is the required solution.