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

this could be newton#'s law of cooling. the cooling curve is exponential

truck y has a greater velocity than truck x.