Problem Statement:
A force vector F = (2i + 3j) N is applied on a particle. The work done is zero when the displacement of the particle x is along 3y, kx = 5. We need to find the value of k.

Solution:

To find the value of k, we can use the formula for work done:

Work done (W) = Force (F) * Displacement (d) * cos(θ)

Here, the work done is given as zero, so we have:

0 = F * d * cos(θ)

Since the displacement is along 3y, we can write the displacement vector as:

d = (0i + 3j)

Now, let's find the angle θ between the force vector F and the displacement vector d.

Angle (θ) = cos^(-1)((F * d) / (|F| * |d|))

Here, F*d = (2 * 0) + (3 * 3) = 9
|F| = √(2^2 + 3^2) = √13
|d| = √(0^2 + 3^2) = 3

Plugging these values into the formula, we get:

θ = cos^(-1)(9 / (3 * √13))

Simplifying further:

θ = cos^(-1)(3 / √13)

Since the work done is zero, cos(θ) must also be zero.

cos(θ) = 0

Let's solve for θ using the value we obtained earlier:

cos^(-1)(3 / √13) = 90°

Therefore, θ = 90°.

We know that cos(90°) = 0, so the value of k is determined by the equation:

kx = 5

Since the displacement is along 3y, x = 0. Substituting this into the equation, we get:

k * 0 = 5

Since any number multiplied by 0 is 0, the equation simplifies to:

0 = 5

This equation is not true for any value of k. Therefore, there is no value of k for which the work done is zero when the displacement is along 3y.