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Jyotinder Singh Grade: 11
        
I am having trouble solving this question:
 
Q. A small body is projected up a rough (u=1/2) inclined plane(theeta = 60 degrees) with a speed of v=10m/s. How far along the plane does it move up before coming to rest? (g=10m/s2).
 
Please solve this using Mechanical energy theorum:
Wexternal + Wnon conservative = (Kf + Uf) – (Ki + Ui)
 
 
My answer is coming out to be 21/2√3. Whereas the answer is given as 20/2√3+1
2 years ago

Answers : (6)

Nicho priyatham
626 Points
										
-umglcos60=mglsin60-(1/2)mv2
-(1/2)(10)(1/2)l= 10(√3/2)l-(1/2)100
so l= 20/(2√3+1)
please approve
 
 
2 years ago
Jyotinder Singh
21 Points
										
Thanks Nicho! That helped me out ^^! I was making a small mistake, I wasn’t multiplying friction by L. Thanks a lot! Is there an upvote system here to choose a correct reply or not? Thanks though!
2 years ago
Jyotinder Singh
21 Points
										
Thanks Nicho! That helped me out ^^! I was making a small mistake, I wasn’t multiplying friction by L. Thanks a lot! Is there an upvote system here to choose a correct reply or not? Thanks though!
2 years ago
Jyotinder Singh
21 Points
										
Thanks Nicho! That helped me out ^^! I was making a small mistake, I wasn’t multiplying friction by L. Thanks a lot! Is there an upvote system here to choose a correct reply or not? Thanks though!
2 years ago
Jyotinder Singh
21 Points
										
R5*TlO#L)_
2 years ago
Nicho priyatham
626 Points
										
ur welcome
 
2 years ago
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