Ahoop of mass 50 kg rolls without slipping. if the center-of-mass of the hoop has a translational speed of 4.0 m/s, the total kinetic energy of the hoop is

a. 0. 20 kj
b. 0.40 kj
c. 1.1 kj
d. 3.9 kj
e. none of these is correct.

total kinetic energy = 800 J

so correct option is e. None of these is correct

Explanation:

given data

hoop  mass = 50 kg

speed = 4.0 m/s

to find out

total kinetic energy of hoop

solution

we know that translational kinetic energy that is  

translational kinetic energy = 0.5×m×v²    ....................1

translational kinetic energy = 0.5×50×4²

translational kinetic energy = 400 J

and rotational kinetic energy  will be

rotational kinetic energy = 0.5 × moment of inertia × (angular speed )²    ............2

here

moment of inertia = mass × R²

and angular speed  =

so rotational kinetic energy = 0.5×m×v²

rotational kinetic energy = 0.5×50×4²

rotational kinetic energy = 400 J

so total kinetic energy = 400 J + 400 J

total kinetic energy = 800 J

so correct option is e. None of these is correct

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge