|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jan 2007 21:09:28 IST
|
|
|
solve - Force acting on a particle is (2i + 3j)N . Work done by this force is zero , when a particle is moved on the line 3y+kx=5 . Hence , find the value of k ?????
|
Umang |
|
|
|
24 Jan 2007 11:04:59 IST
|
|
|
Let us consider two points on the given line be P(0,5/3) and Q(1, (5-k)/3) This can be represented by radius vector r= i -kj Since work done is zero Hence F dot r=0 (2i+3j).(i-kj)=0 2-3k=0 k=2/3
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jan 2007 21:05:21 IST
|
|
|
Hi yahiyafirdous , I am not able to understand how u got radius vector = i-kj . also , the options i have for the value of k are 2 , 4 , 6 & 8
|
Umang |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
24 Jan 2007 21:29:54 IST
|
|
|
4 might be the ans
|
ADARSH
NITK Surathkal
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
25 Jan 2007 20:45:58 IST
|
|
|
Hi , the correct answer is k=2 , and i got it . work done = F.dx , where dx is the distance b/w final and initial points . let's take two points (0,5/3) and (5/k,0) which lie on the given line Then , dx = 5/ki + 5/3j hence , F.dx=0 (bcoz W=0) i.e. 10/k-5=0 k=2
|
Umang |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
