A block of mass 0.18 kg is attached to a spring of force-constant $2 N/m.$ The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in $m/s$ is $V = N/10.$ Then N is

Step 1: Identify the given parameters.

The physical constants and initial conditions of the system are given.

$$m = 0.18 \text{ kg}$$ $$k = 2 \text{ N/m}$$ $$\mu = 0.1$$ $$d = 0.06 \text{ m}$$ $$g = 10 \text{ m/s}^{2}$$

Step 2: Apply the work-energy theorem.

The initial kinetic energy of the block is converted into the potential energy stored in the spring and the work done against friction.

$$\frac{1}{2}mV^{2} = \frac{1}{2}kd^{2} + \mu mgd$$

Step 3: Solve for the initial velocity $V$.

Rearrange the equation to express $V^{2}$ in terms of the known variables.

$$V^{2} = \frac{k}{m}d^{2} + 2\mu gd$$

Substitute the given values into the equation.

$$V^{2} = \frac{2}{0.18}(0.06)^{2} + 2(0.1)(10)(0.06)$$ $$V^{2} = 0.04 + 0.12$$ $$V^{2} = 0.16$$ $$V = 0.4 \text{ m/s}$$

Step 4: Find the value of $N$.

Equate the calculated velocity to the given expression $V = \frac{N}{10}$.

$$\frac{N}{10} = 0.4$$ $$N = 4$$

Answer: $N = 4$