A120 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. how much work must be done on the hoop to stop it?

W = -5.808J

Explanation:

Given:

Mass of hoop,m=120kg

Centre of mass speed of hoop,v= 0.220m/s

Rotational inertia,I= mr^2

I=120r^2

Kinetic energy of hoop= Linear kinetic energy + Rotational kinetic energy

K= (1/2mv^2) + (1/2Iw^2)

But w= (v/r)^2

K= (1/2mv^2) + (1/2×120r^2(v/r)^2)

K = (1/2×120×0.220^2) + (1/2×120×0.220^2)

K= 5.808J

To stop the hoop,its final kinetic energy must be zero,

Workdone = Kfinal-Kinitial = 0 - 5.808

W= -5.808J

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

its a 5.5 n

glad that