Ahoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 25° below the horizontal. what is the rotational kinetic energy of the hoop after it has rolled 3.0 m down as measured along the surface of the ramp?

Explanation:

given,

mass of hoop = 2.75 Kg

horizontal speed = 4.5 m/s

angle with horizontal = 25°

m g l sin θ = m ( v² - u² )

g l sin θ = ( v² - u² )

9.8 x 3 x sin 25° = ( v² - 4.5² )

12.425 = v² - 20.25

v² = 32.675

v = 5.72 m/s

Rotational

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.