Questions: A block of mass 1.9 kg is attached to a horizontal spring that has a force constant 900 N/m as shown in the figure below. The spring is compressed 2.0 cm and is then released from rest. (a) A constant friction force of 4.1 N retards the block's motion from the moment it is released. How much is the spring compressed when the speed of the block is a maximum. cm (b) What is the maximum speed? cm/s

Solution Steps

• Mass of the block, \( m = 1.9 \, \text{kg} \)
• Spring constant, \( k = 900 \, \text{N/m} \)
• Initial compression of the spring, \( x_0 = 2.0 \, \text{cm} = 0.02 \, \text{m} \)
• Friction force, \( f = 4.1 \, \text{N} \)

The initial potential energy stored in the spring is given by: \[ E_{\text{spring}} = \frac{1}{2} k x_0^2 \] \[ E_{\text{spring}} = \frac{1}{2} \times 900 \, \text{N/m} \times (0.02 \, \text{m})^2 \] \[ E_{\text{spring}} = \frac{1}{2} \times 900 \times 0.0004 \] \[ E_{\text{spring}} = 0.18 \, \text{J} \]

The work done by friction when the block moves a distance \( x \) is: \[ W_{\text{friction}} = f \cdot x \] \[ W_{\text{friction}} = 4.1 \, \text{N} \cdot x \]

At maximum speed, the kinetic energy is at its peak, and the potential energy in the spring is partially converted to kinetic energy and work done against friction. The energy conservation equation is: \[ \frac{1}{2} k x_0^2 = \frac{1}{2} m v_{\text{max}}^2 + f \cdot x \]

At maximum speed, the spring is neither fully compressed nor fully extended, so let \( x \) be the compression at maximum speed. The potential energy in the spring at this point is: \[ E_{\text{spring, max}} = \frac{1}{2} k x^2 \]

The energy conservation equation becomes: \[ \frac{1}{2} k x_0^2 = \frac{1}{2} k x^2 + f \cdot x \] \[ 0.18 = \frac{1}{2} \times 900 \times x^2 + 4.1 \times x \] \[ 0.18 = 450 x^2 + 4.1 x \]

Rearrange the equation to form a standard quadratic equation: \[ 450 x^2 + 4.1 x - 0.18 = 0 \]

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 450, \, b = 4.1, \, c = -0.18 \] \[ x = \frac{-4.1 \pm \sqrt{(4.1)^2 - 4 \cdot 450 \cdot (-0.18)}}{2 \cdot 450} \] \[ x = \frac{-4.1 \pm \sqrt{16.81 + 324}}{900} \] \[ x = \frac{-4.1 \pm \sqrt{340.81}}{900} \] \[ x = \frac{-4.1 \pm 18.46}{900} \]

Taking the positive root: \[ x = \frac{14.36}{900} \] \[ x \approx 0.01595 \, \text{m} \] \[ x \approx 1.595 \, \text{cm} \]

Final Answer