Question

A 140 kg hoop rolls along a horizontal floor so that the hoop’s center of mass has a speed of 0.125 m/s. How much work must be done on the hoop to stop it?​

Answer 1

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Answer 2

Given:

Mass of hoop rolls = 140kg.

Speed = 0.125 m/s.

To Find:

Work done on the hoop to stop it.

Solution:

Let the radius of the hoop be R,

Then, the Inertia of the hoop is,

I = mass×R².

I = 140×R².

Total work done can be calculated as,

W = change in kinetic energy.

W = Linear kinetic energy + Rotational kinetic energy.

W= 1/2 mV² + 1/2 Iω².

W= 1/2 mV² + 1/2 I(V²/R²).

W= 1/2 mV² + 1/2 (MR²)(V²/R²).

W = 5.6 Joules.

Hence, the work done on the hoop to stop it is 5.6 Joules.