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I = 1/2MR^2 is rolling with a translational speed v along a horizontal surface without slipping. The work required to stop the disk is A. 1/2 mv^2 B. 1/2mv^2 + 1/2Iw^2 C. 1/2Iw^2 D. 1/2mv^2 - 1/2Iw^2
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I = 1/2MR^2 is rolling with a translational speed v along a horizontal surface without slipping. The work required to stop the disk is

A. 1/2 mv^2
B. 1/2mv^2 + 1/2Iw^2
C. 1/2Iw^2
D. 1/2mv^2 - 1/2Iw^2

User Jon Onstott
by
8.8k points

1 Answer

2 votes

Answer:

B. 1/2mv^2 + 1/2Iw^2

Step-by-step explanation:

User Sebnukem
by
7.9k points
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