Let's break the problem into **two main steps**: 1. **Set up the differential equation of motion.** 2. **Solve the for \( x \) at \( t = 2 \) s.** --- ## **Step 1: Set Up the Equation of Motion** - **Forces on the block:** - Gravity down the incline: \( 30g \sin(20^\circ) \) - Spring force: \( -kx \) - Damping force: \( -c\dot{x} \) - **Pulley consideration:** - The tension in the cable also accelerates the hanging mass (30 kg) and rotates the pulley (radius \( R = .12 \) m, moment of inertia \( I = .03 \) kg·m²). Let's derive the equation: **Total mass:** - Block: \( m = 30 \) kg - Hanging mass: \( m = 30 \) kg - Pulley: effective mass via \( I/R^2 = .03/.12^2 = 2.083 \) kg **Total effective mass:** \[ m_{\text{total}} = 30 + 30 + 2.083 = 62.083 \text{ kg} \] **Equation (using \( x \) as positive downward for hanging mass):** \[ m_{\text{total}} \ddot{x} + c\dot{x} + kx = 30g \] **Plug in the numbers:** - \( c = 250 \) N·s/m - \( k = 800 \) N/m - \( g = 9.81 \) m/s² \[ 62.083\ddot{x} + 250\dot{x} + 800x = 30 \times 9.81 \] \[ 62.083\ddot{x} + 250\dot{x} + 800x = 294.3 \] **Explanation:** We set up the differential equation by summing all forces and converting rotational inertia of the pulley to an equivalent mass. This creates a standard second-order ODE with constant coefficients. --- ## **Step 2: Solve for \( x \) at \( t = 2 \) s** ### 1. **Find the homogeneous and particular solution** - The ODE is: \[ 62.083\ddot{x} + 250\dot{x} + 800x = 294.3 \] - Characteristic equation: \[ 62.083r^2 + 250r + 800 = \] Solve for \( r \): \[ r = \frac{-250 \pm \sqrt{250^2 - 4 \cdot 62.083 \cdot 800}}{2 \cdot 62.083} \] \[ = \frac{-250 \pm \sqrt{62500 - 198665.6}}{124.166} \] \[ = \frac{-250 \pm \sqrt{-136165.6}}{124.166} \] \[ = \frac{-250 \pm 369.1i}{124.166} \] \[ = -2.014 \pm 2.973i \] - **Homogeneous solution:** \[ x_h(t) = e^{-2.014t} \left[ A \cos(2.973t) + B \sin(2.973t) \right] \] - **Particular solution:** For constant input, steady-state is \[ x_p = \frac{294.3}{800} = .368 \] - **General solution:** \[ x(t) = x_p + x_h(t) = .368 + e^{-2.014t} \left[ A \cos(2.973t) + B \sin(2.973t) \right] \] ### 2. **Apply initial conditions** (\( x() = \), \( \dot{x}() = 1 \) m/s): - At \( t = \): \[ x() = .368 + A = \implies A = -.368 \] - Derivative: \[ \dot{x}(t) = e^{-2.014t} \left[ -2.014A \cos(2.973t) - 2.014B \sin(2.973t) -2.973A \sin(2.973t) + 2.973B \cos(2.973t) \right] \] At \( t = \): \[ \dot{x}() = -2.014A + 2.973B = 1 \] \[ -2.014(-.368) + 2.973B = 1 \] \[ .741 + 2.973B = 1 \] \[ 2.973B = .259 \implies B = .087 \] ### 3. **Write the full solution:** \[ x(t) = .368 + e^{-2.014t} \left[ -.368 \cos(2.973t) + .087 \sin(2.973t) \right] \] ### 4. **Plug in \( t = 2 \) s:** Compute each term: - \( e^{-2.014 \times 2} = e^{-4.028} \approx .0178 \) - \( \cos(2.973 \times 2) = \cos(5.946) \approx .943 \) - \( \sin(5.946) \approx -.332 \) So, \[ x(2) = .368 + .0178 [ -.368 \times .943 + .087 \times (-.332) ] \] \[ = .368 + .0178 [ -.3468 - .0288 ] \] \[ = .368 + .0178 \times (-.3756) \] \[ = .368 - .00669 \] \[ = .361 \text{ m} \] --- ## **Final Answer** ### **The value of \( x \) at \( t = 2 \) s is \(\boxed{.36~\mathrm{m}}\) (rounded to two decimal places).**