Let's break down the problem step by step. ### **Step 1: Find the Speed after 89 cm Down the Plane** #### **Given Data:** - \( F_{yb} = 56 \) N (horizontal force) - \( m_b = 6.3 \) kg (block mass) - \( m_c = 4.2 \) kg (cylinder mass) - \( r = 0.4 \) m (cylinder radius) - \( k = 155 \) N/m (spring constant) - \( \theta = 30^\circ \) (incline angle) - \( \mu_s = 0.66 \) - \( \mu_k = 0.36 \) - \( d = 0.89 \) m (distance down the plane) #### **Step 1.1: Work-Energy Principle** The initial energy is zero (everything starts at rest, spring is unstretched). **Work done by all forces = Change in kinetic energy + Change in spring potential energy** - **Work by pull force (\(F_{yb}\)):** - The force is horizontal, but the block moves down the incline. - Component of force along the incline, \( F_{\text{along}} = F_{yb} \cos\theta \) - Work by \( F_{yb} \): \( W_F = F_{yb} \cos\theta \cdot d \) - **Work by gravity:** - Component down the plane: \( (m_b + m_c)g \sin\theta \) - Work: \( W_g = (m_b + m_c)g \sin\theta \cdot d \) - **Work by friction:** - Opposes motion, kinetic friction: \( \mu_k (m_b + m_c)g \cos\theta \) - Work: \( W_{fric} = -\mu_k (m_b + m_c)g \cos\theta \cdot d \) - **Spring force:** - At 0.89 m, the spring is stretched \( x = d \). - Potential energy stored: \( U_s = \frac{1}{2} k x^2 \) #### **Step 1.2: Kinetic Energy** - Block: \( \frac{1}{2} m_b v^2 \) - Cylinder (translational + rotational): - Translational: \( \frac{1}{2} m_c v^2 \) - Rotational: \( \frac{1}{2} I \omega^2 \), for solid cylinder \( I = \frac{1}{2} m_c r^2 \), and \( v = r\omega \Rightarrow \omega = v/r \) - So, \( \frac{1}{2} \cdot \frac{1}{2} m_c r^2 \cdot (v/r)^2 = \frac{1}{4} m_c v^2 \) - **Total KE:** \( \frac{1}{2} m_b v^2 + \frac{1}{2} m_c v^2 + \frac{1}{4} m_c v^2 = \frac{1}{2} m_b v^2 + \frac{3}{4} m_c v^2 \) #### **Step 1.3: Energy Equation** \[ \text{Net Work} = \Delta KE + \Delta U_s \] \[ F_{yb} \cos\theta \cdot d + (m_b + m_c)g \sin\theta \cdot d - \mu_k (m_b + m_c)g \cos\theta \cdot d = \left(\frac{1}{2} m_b + \frac{3}{4} m_c\right)v^2 + \frac{1}{2} k d^2 \] #### **Plug in Numbers:** - \( F_{yb} = 56 \) N - \( \cos\theta = \cos(30^\circ) = 0.866 \) - \( \sin\theta = 0.5 \) - \( g = 9.8 \) m/s² - \( m_b + m_c = 10.5 \) kg - \( \mu_k = 0.36 \) - \( k = 155 \) N/m - \( d = 0.89 \) m 1. \( F_{yb} \cos\theta \cdot d = 56 \cdot 0.866 \cdot 0.89 = 43.5 \) J 2. \( (m_b+m_c)g\sin\theta \cdot d = 10.5 \cdot 9.8 \cdot 0.5 \cdot 0.89 = 45.8 \) J 3. \( \mu_k (m_b+m_c)g\cos\theta \cdot d = 0.36 \cdot 10.5 \cdot 9.8 \cdot 0.866 \cdot 0.89 = 28.3 \) J 4. \( \frac{1}{2} k d^2 = 0.5 \cdot 155 \cdot (0.89)^2 = 61.3 \) J 5. \( \frac{1}{2} m_b + \frac{3}{4} m_c = 0.5 \cdot 6.3 + 0.75 \cdot 4.2 = 3.15 + 3.15 = 6.3 \) Sum up the work: \[ \text{Net Work} = 43.5 + 45.8 - 28.3 = 61.0 \text{ J} \] So, \[ 61.0 = 6.3 v^2 + 61.3 \] \[ 6.3 v^2 = 61.0 - 61.3 = -0.3 \] \[ v^2 = -0.3 / 6.3 = -0.0476 \] Since the value is negative, let's check the arithmetic: - \( F_{yb} \cos\theta \cdot d = 56 \times 0.866 \times 0.89 = 43.3 \) - \( (m_b+m_c)g\sin\theta \cdot d = 10.5 \times 9.8 \times 0.5 \times 0.89 = 45.8 \) - \( \mu_k (m_b+m_c)g\cos\theta \cdot d = 0.36 \times 10.5 \times 9.8 \times 0.866 \times 0.89 \approx 28.3 \) - \( \frac{1}{2} k d^2 = 0.5 \times 155 \times 0.7921 = 61.6 \) - \( \frac{1}{2} m_b + \frac{3}{4} m_c = 3.15 + 3.15 = 6.3 \) \[ \text{Net Work} = 43.3 + 45.8 - 28.3 = 60.8 \] \[ 6.3 v^2 + 61.6 = 60.8 \] \[ 6.3 v^2 = 60.8 - 61.6 = -0.8 \] \[ v^2 = -0.8 / 6.3 = -0.127 \] A negative value indicates that all the input energy is absorbed by the spring and friction, and the system cannot reach this stretch (it never gets to 0.89 m before stopping). **So the speed is 0 m/s (they stop before reaching 0.89 m).** --- ### **Step 2: How Far Have You Pulled When Everything Stops?** Set final kinetic energy to zero: \[ F_{yb} \cos\theta \cdot d + (m_b + m_c)g \sin\theta \cdot d - \mu_k (m_b + m_c)g \cos\theta \cdot d = \frac{1}{2} k d^2 \] \[ [F_{yb} \cos\theta + (m_b+m_c)g\sin\theta - \mu_k (m_b+m_c)g\cos\theta]d = \frac{1}{2}k d^2 \] \[ a d = \frac{1}{2} k d^2 \] \[ a = \frac{1}{2} k d \] \[ d = \frac{2a}{k} \] Where \[ a = F_{yb} \cos\theta + (m_b+m_c)g\sin\theta - \mu_k (m_b+m_c)g\cos\theta \] \[ a = 43.3 + 45.8 - 28.3 = 60.8 \] \[ d = \frac{2 \times 60.8}{155} = \frac{121.6}{155} \approx 0.785 \text{ m} \] --- ## **Final Answers** ### **Step 1: Speed after 89 cm down the plane** **Answer:** \( v = 0 \) m/s **Explanation:** The block and cylinder stop before reaching 89 cm because all input energy is absorbed by the spring and friction. --- ### **Step 2: How far before everything stops?** **Answer:** \( d_{stop} = 0.785 \) m **Explanation:** This is the distance at which all the work done is absorbed by friction and the stretched spring, so the objects come to rest. --- #### Let me know if you need further clarification or calculations!

The problem uses the work-energy principle, which states that the net work done on the system equals the change in its kinetic energy plus the change in potential energy stored in the spring. To find the speed after moving a certain distance, we calculate the work done by applied force, gravity, and friction over that distance and compare it to the total kinetic and elastic potential energy gained. To determine the stopping distance, we set the kinetic energy to zero and equate the work done by the forces to the energy absorbed by friction and the spring, solving for the distance where motion ceases.