Calculate Initial Velocity of Block Attached to Spring with Friction

To find the value of $N$, we apply the Work-Energy Theorem to the block from the instant right after the impulse is delivered until it comes to rest for the first time.

1. Given Data:

• Mass of the block, $m = 0.18\text{ kg}$
• Force constant of the spring, $k = 2\text{ N/m}$
• Coefficient of kinetic friction, $\mu = 0.1$
• Distance moved by the block, $x = 0.06\text{ m}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$
• Initial speed after impulse, $V = \frac{N}{10}\text{ m/s}$
• Final speed when it comes to rest, $v_f = 0\text{ m/s}$

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2. Work-Energy Theorem:

The Work-Energy Theorem states that the net work done on the block equals its change in kinetic energy:
$W_{\text{net}} = \Delta K$

$W_{\text{spring}} + W_{\text{friction}} = K_f - K_i$

Where:

• Work done by the spring force:
  $W_{\text{spring}} = -\frac{1}{2} k x^2$

• Work done by the frictional force:
  $W_{\text{friction}} = - f_k \cdot x = -\mu m g x$

• Initial kinetic energy:
  $K_i = \frac{1}{2} m V^2$

• Final kinetic energy:
  $K_f = 0$

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3. Setting up the Equation:

Substituting these expressions into the work-energy relation:
$-\frac{1}{2} k x^2 - \mu m g x = 0 - \frac{1}{2} m V^2$

Multiplying both sides by $-2$:
$m V^2 = k x^2 + 2 \mu m g x$

Dividing by $m$:
$V^2 = \frac{k}{m} x^2 + 2 \mu g x$

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4. Numerical Calculation:

Substitute the given values into the equation:

1. Spring term:
   $\frac{k}{m} x^2 = \frac{2}{0.18} \times (0.06)^2 = \frac{2}{0.18} \times 0.0036 = \frac{0.0072}{0.18} = 0.04\text{ m}^2/\text{s}^2$

2. Friction term:
   $2 \mu g x = 2 \times 0.1 \times 10 \times 0.06 = 0.12\text{ m}^2/\text{s}^2$

Adding the two components together:
$V^2 = 0.04 + 0.12 = 0.16\text{ m}^2/\text{s}^2$

Taking the square root:
$V = \sqrt{0.16} = 0.4\text{ m/s}$

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5. Finding $N$:

We are given that $V = \frac{N}{10}$. Equating the two values:
$0.4 = \frac{N}{10} \implies N = 4$