Let's solve each step by step--- ## () **Find time needed for block to reach velocity of m/s** ###Step 1 Write the equation motion** The force on the is: \F_{\text{}} = Ftext{ap}} - (_c + F + F_s\] Given: \( F_{\{applied = 8) N \( F_c 1 \ N - F_v = || \) N- \( F = \frac4}{|v|+1} \) N - \( m = 1 \) kg So, \[ m \frac{dv}{dt} = 8 - \left(1 + v + \frac{4}{v+1}\right) \] \[ \frac{dv}{dt} = 8 - 1 - v - \frac{4}{v+1} \] \[ \frac{dv}{dt} = 7 - v - \frac{4}{v+1} \] **Explanation:** We have written the net force acting on the block by subtracting all (velocity-dependent) friction forces from the applied force. --- ### **Step 2: Set up and solve the differential equation** We want to find the time to reach \( v = 1 \) m/s, starting from rest (\( v = \)). \[ \frac{dv}{dt} = 7 - v - \frac{4}{v+1} \] \[ dt = \frac{dv}{7 - v - \frac{4}{v+1}} \] Let's simplify the denominator: \[ 7 - v - \frac{4}{v+1} = \frac{7(v+1) - v(v+1) - 4}{v+1} = \frac{7v + 7 - v^2 - v - 4}{v+1} = \frac{-v^2 + 6v + 3}{v+1} \] So, \[ dt = \frac{v+1}{-v^2 + 6v + 3} dv \] The time to reach \( v = 1 \) is: \[ t = \int_{v=}^{v=1} \frac{v+1}{-v^2 + 6v + 3} dv \] **Explanation:** We set up the integral to solve for the time by separating variables and simplifying the expression. --- #### **Final Answer (a):** The time required is: \[ \boxed{ t = \int_{}^{1} \frac{v+1}{-v^2 + 6v + 3}\, dv } \] This can be solved numerically if an explicit answer is needed. --- ## (b) **What is the limiting velocity that the block approaches after a long time?** ### **Step 1: Set acceleration to zero for steady state** At limiting velocity (\( v_\infty \)), acceleration is zero (\( \frac{dv}{dt} = \)): \[ = 8 - (1 + v_\infty + \frac{4}{v_\infty+1}) \] \[ = 7 - v_\infty - \frac{4}{v_\infty+1} \] \[ v_\infty + \frac{4}{v_\infty+1} = 7 \] **Explanation:** We set the net force to zero to find the terminal (limiting) velocity. --- ### **Step 2: Solve for \( v_\infty \) algebraically** Multiply both sides by \( v_\infty + 1 \): \[ v_\infty (v_\infty + 1) + 4 = 7 (v_\infty + 1) \] \[ v_\infty^2 + v_\infty + 4 = 7v_\infty + 7 \] \[ v_\infty^2 - 6v_\infty - 3 = \] Solve the quadratic: \[ v_\infty = \frac{6 \pm \sqrt{36 + 12}}{2} = \frac{6 \pm \sqrt{48}}{2} = \frac{6 \pm 4\sqrt{3}}{2} = 3 \pm 2\sqrt{3} \] Since velocity must be positive, \[ v_\infty = 3 + 2\sqrt{3} \approx 6.464 \] **Explanation:** We solved the quadratic equation for the steady-state velocity, keeping only the positive root. --- #### **Final Answer (b):** \[ \boxed{ v_\infty = 3 + 2\sqrt{3} \approx 6.464~\text{m/s} } \]

# Friction Model for Block Movement We will solve the problem of a block moving on a surface with friction in two steps: finding the time needed to reach a specified velocity and determining the limiting velocity. ## (a) Find Time Needed for Block to Reach Velocity of 1 m/s ### Step 1: Write the Equation of Motion The net force acting on the block is given by: \[ F_{\text{net}} = F_{\text{applied}} - (F_c + F_v + F_s) \] Where: - \( F_{\text{applied}} = 8 \, \text{N} \) - \( F_c = 1 \, \text{N} \) (Coulomb force) - \( F_v = |v| \, \text{N} \) (Viscous force) - \( F_s = \frac{4}{|v| + 1} \, \text{N} \) (Stribeck force) - Mass \( m = 1 \, \text{kg} \) The equation of motion becomes: \[ m \frac{dv}{dt} = 8 - \left(1 + v + \frac{4}{v + 1}\right) \] This simplifies to: \[ \frac{dv}{dt} = 7 - v - \frac{4}{v + 1} \] **Explanation:** We derived the motion equation by subtracting the total frictional forces from the applied force acting on the block. --- ### Step 2: Set Up and Solve the Differential Equation To find the time to reach \( v = 1 \, \text{m/s} \) from rest, we rearrange the equation: \[ dt = \frac{dv}{7 - v - \frac{4}{v + 1}} \] Next, we simplify the denominator: \[ 7 - v - \frac{4}{v + 1} = \frac{7(v + 1) - v(v + 1) - 4}{v + 1} = \frac{-v^2 + 6v + 3}{v + 1} \] Thus, the equation becomes: \[ dt = \frac{v + 1}{-v^2 + 6v + 3} dv \] The total time to reach \( v = 1 \) is: \[ t = \int_{0}^{1} \frac{v + 1}{-v^2 + 6v + 3} \, dv \] **Explanation:** We set up the integral to compute the time needed by separating the variables and simplifying the expression for integration. --- #### Final Answer (a): \[ \boxed{ t = \int_{0}^{1} \frac{v + 1}{-v^2 + 6v + 3} \, dv } \] --- ## (b) What is the Limiting Velocity that the Block Approaches After a Long Time? ### Step 1: Set Acceleration to Zero for Steady State At limiting velocity (\( v_\infty \)), the acceleration equals zero: \[ 0 = 8 - (1 + v_\infty + \frac{4}{v_\infty + 1}) \] This leads to: \[ v_\infty + \frac{4}{v_\infty + 1} = 7 \] **Explanation:** Setting the net force to zero allows us to find the terminal velocity where the forces balance out. --- ### Step 2: Solve for \( v_\infty \) Algebraically Multiply both sides by \( v_\infty + 1 \): \[ v_\infty (v_\infty + 1) + 4 = 7(v_\infty + 1) \] Simplifying gives: \[ v_\infty^2 + v_\infty + 4 = 7v_\infty + 7 \] \[ v_\infty^2 - 6v_\infty - 3 = 0 \] Using the quadratic formula: \[ v_\infty = \frac{6 \pm \sqrt{36 + 12}}{2} = \frac{6 \pm \sqrt{48}}{2} = \frac{6 \pm 4\sqrt{3}}{2} = 3 \pm 2\sqrt{3} \] Since velocity must be positive, we take: \[ v_\infty = 3 + 2\sqrt{3} \approx 6.464 \, \text{m/s} \] **Explanation:** We solved the quadratic equation for the steady-state velocity, discarding the negative root to find the physical solution. --- #### Final Answer (b): \[ \boxed{ v_\infty = 3 + 2\sqrt{3} \approx 6.464 \, \text{m/s} } \]