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Mar 24, 2021 at 4:11 vote accept Yash Agrawal
Mar 24, 2021 at 4:11 answer added Yash Agrawal timeline score: 0
Mar 15, 2021 at 13:05 vote accept Yash Agrawal
Mar 24, 2021 at 4:11
Mar 15, 2021 at 0:57 answer added Claudio Saspinski timeline score: 0
S Mar 14, 2021 at 23:59 history suggested kbakshi314 CC BY-SA 4.0
grammatical edits and formatting changes and added tags
Mar 14, 2021 at 23:21 review Suggested edits
S Mar 14, 2021 at 23:59
Mar 14, 2021 at 23:04 history edited Qmechanic CC BY-SA 4.0
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Mar 14, 2021 at 19:05 answer added AnyaC timeline score: 0
Mar 14, 2021 at 14:38 comment added Yash Agrawal that's what i am also saying
Mar 14, 2021 at 14:37 comment added Brian I mean if you denote work as $ \vec{F} \cdot ds$ , then the earth is not really be pushed down by any distance so the work done on earth must be zero
Mar 14, 2021 at 14:36 comment added Yash Agrawal For example there is a small block placed on the ground and then it is raised by the height 'h', then work done by gravitational force is mgh but same gravitational force is acting on the earth but earth's mass is tremendously greater than block so there would be negligible work done by gravitational force on the earth , so mgh is the the only work done on the earth + block system ; that's what I mean ;Is it correct ?
Mar 14, 2021 at 14:26 comment added Brian >For example in case of gravitational potential energy ,we only consider work done on small body as there is negligible work done on the earth. This statement is not very clear for me
Mar 14, 2021 at 14:23 answer added Brian timeline score: 0
Mar 14, 2021 at 14:09 history asked Yash Agrawal CC BY-SA 4.0