Question

A block is fastened to a horizontal spring. The block is pulled to a distance \(x=10\ cm\) from its equilibrium position (at \(x=0\) ) on a frictionless surface from rest. The energy of the block at \(x=5\) \(cm\) is \(025 J\). The spring constant of the spring is _____\(Nm ^{-1}\).

Hint:

Remember the formulas for potential and kinetic energy. Applying the principle of conservation of energy can help solve problems involving spring systems.

Answer 1

Correct Answer: 67

Step 1: Analyze the Initial Energy

When the block is pulled to \(x_0 = 10 \, \text{cm} = 0.1 \, \text{m}\), all the energy is stored as potential energy in the spring:

\[ U_i = \frac{1}{2} k x_0^2 \]

where \(k\) is the spring constant. The initial kinetic energy is zero because the block is at rest.

Step 2: Analyze the Energy at \(x = 5 \, \text{cm}\)

When the block is at \(x = 5 \, \text{cm} = 0.05 \, \text{m} = x_0 / 2\), the total energy is the sum of the potential energy and kinetic energy:

\[ U_f = \frac{1}{2} k \left(\frac{x_0}{2}\right)^2 = \frac{1}{8} k x_0^2 \]

\[ K_f = 0.25 \, \text{J}. \]

Total energy at this point will be the sum of potential and kinetic energy.

Step 3: Apply Conservation of Energy

Since the surface is frictionless, the total mechanical energy is conserved. Therefore, the initial energy equals the final energy:

\[ U_i + K_i = U_f + K_f \] \[ \frac{1}{2} k x_0^2 + 0 = \frac{1}{8} k x_0^2 + 0.25 \] \[ \frac{1}{2} k x_0^2 - \frac{1}{8} k x_0^2 = 0.25 \] \[ \frac{3}{8} k x_0^2 = 0.25 \]

Step 4: Solve for the Spring Constant (\(k\))

\[ k = \frac{0.25 \times 8}{3 x_0^2} = \frac{2}{3 x_0^2} \]

Substitute \(x_0 = 0.1 \, \text{m}\):

\[ k = \frac{2}{3 (0.1)^2} = \frac{2}{3 \times 0.01} = \frac{2}{0.03} = \frac{200}{3} \approx 66.67 \, \text{N/m}. \]

Rounding to the nearest integer gives \(67 \, \text{N/m}\).

Conclusion: The spring constant is approximately \(67 \, \text{N/m}\).