IIT JEE , AIEEE Forum - Physics , Mechanics

umang

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 ?????

yahiyafirdous

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

umang

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

KAB

4 might be the ans

umang

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

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