Answer :
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,
[tex]K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
I is the moment of inertia, [tex]I=mr^2[/tex]
Since, [tex]\omega=\dfrac{v}{r}[/tex]
[tex]K_i=mv^2[/tex]
[tex]K_i=96\times (0.24)^2=5.529\ J[/tex]
Finally it stops, so the final energy of the hoop will be, [tex]K_f=0[/tex]
The work done by the hoop is equal to the change in kinetic energy as :
[tex]W=K_f-K_i[/tex]
W = -5.529 Joules
So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.