Question

A 110 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

Answer 1

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

Answer 2

The amount of work that must be done to stop a hoop of mass 110 kg moving with s speed of 0.220 m/s is 2.662 J

Work can be defined as the ability or the capacity to perform work. Work is also the measure of energy of a body. The S.I unit of work is Joules (J)

• From the question,                                                                                     The kinetic energy of the hoop = The energy required to stop the hoop.

Formular for kinetic energy of the hoop is

EK = mv²/2.................... Equation 1

Where EK = Kinetic energy of the hoop, m = mass of the hoop, v = velocity of the hoop.

Given: m = 110kg, v = 0.220 m/s.

Substitute these values into equation 1

EK = 110(0.22)²/2

Ek = 2.662 J.

 

Therefore, The amount of work that must be done on the hoop to stop it is

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