Question

A force F=6i+2j-3k acts on a particle and produces a displacement ofs=2i-3j-xk . if the work done is zero the value of x is

Answer 1

to work done is zero, Force and displacement to be perpendicular of each other if both are non zero.
Hence 6(2) + 2(-3) +(-3)(-x) = 0
as, a1a2 + b1b2 + c1c2 = 0 for mutually perpendicular
hence 12 -6 +3x = 0
x = -2

Hope you get it. :)

Answer 2

Answer:

x = -2

Explanation:

Given that,

Force acting on the particle, $F=6i+2j-3k$

Displacement due to this force, $x=2i-3j-xk$

Work done, W = 0

We know that the dot product of force and displacement is called the work done. Mathematically, it is given by :

$W=F.x$

$(6i+2j-3k).(2i-3j-xk)=0$

Since, i.i = j.j = k.k = 1

x = -2

So, the value of x is -2. Hence, this is the required solution.