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 d one on the hoop to stop it
1 answer:
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
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