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2 \times 3
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Comments
Since normal( Between block and wedge) is not a dissipative force hence net work done by it on the system is zero.
Now the lower wedge is getting velocity because of the normal force hence it is doing positive work on the wedge , since net work is zero it implies normal force does negative work on the block.
Ans) Negative
Another way to analyse the FBD of the block
The normal force is making an angle \theta with the positive y-axis and the block's displacement has two components one perpendicular to block and one along the -x axis. So work done in first component is zero and work done in second component is -Nxsin(\theta) ( The angle between normal and displacement(x) is 90+\theta)
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Since normal( Between block and wedge) is not a dissipative force hence net work done by it on the system is zero.
Now the lower wedge is getting velocity because of the normal force hence it is doing positive work on the wedge , since net work is zero it implies normal force does negative work on the block.
Ans) Negative
Another way to analyse the FBD of the block
The normal force is making an angle \theta with the positive y-axis and the block's displacement has two components one perpendicular to block and one along the -x axis. So work done in first component is zero and work done in second component is -Nxsin(\theta) ( The angle between normal and displacement(x) is 90+\theta)
It is clear work done is negative.
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Bhiya Can U give me ur Email ID ?
Please Respond.
Best of luck Bhaiya for Sunday. I am sure U will get AIR 1. ⌣¨
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If by chance I would get a rank just under 100 I would be over the moon and you are talking about 1 Ha Ha Ha.
Yes You are Right Here..!! Kudos..!!
W=F.S when force is constant.
As clearly seen from diagram,angle b/w force and displacement is 90.
Hence, work done is 0
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@Akhil Bansal I am sorry but the angle between force and displacement here is not 90..!!
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Can you see this ?