Let's solve each part step by step, with explanations after each part. --- ### **a) Nana’s Acceleration When the Rope Goes Taut** #### **Step 1: Draw Free Body Diagrams & Identify Forces** - **Nana (on slope):** - Gravity: \(mg\) (downwards, resolve into parallel and perpendicular components) - Friction: \(f_k = \mu_k N\) (up the slope) - Tension: \(T\) (up the slope) - **Popo (hanging vertically):** - Gravity: \(Mg\) (downwards) - Tension: \(T\) (upwards) #### **Step 2: Set Up Equations** - For Nana (along the slope): \[ m a = m g \sin\theta - f_k - T \] But, since she is being pulled up the slope by the rope, the direction of tension is *down the slope* (same as the direction of motion after Popo falls). So, \(a =\) acceleration **down** the slope. Friction force: \(f_k = \mu_k N = \mu_k m g \cos\theta\) Thus: \[ m a = m g \sin\theta - \mu_k m g \cos\theta + T \] (Tension **adds** because it pulls Nana down the slope.) - For Popo (vertical motion): \[ M a = M g - T \] \[ T = M g - M a \] #### **Step 3: Combine Equations and Solve for a** Substitute \(T\) from Popo's equation into Nana's: \[ m a = m g \sin\theta - \mu_k m g \cos\theta + (M g - M a) \] \[ m a + M a = m g \sin\theta - \mu_k m g \cos\theta + M g \] \[ (m + M) a = m g \sin\theta - \mu_k m g \cos\theta + M g \] \[ a = \frac{m g \sin\theta - \mu_k m g \cos\theta + M g}{m + M} \] Plug in values: - \(m = 60\) kg - \(M = 80\) kg - \(g = 9.8\) m/s² - \(\mu_k = 0.7\) - \(\theta = 40^\circ\) Calculate: - \(\sin 40^\circ \approx 0.643\) - \(\cos 40^\circ \approx 0.766\) \[ a = \frac{60 \times 9.8 \times 0.643 - 0.7 \times 60 \times 9.8 \times 0.766 + 80 \times 9.8}{60 + 80} \] \[ = \frac{60 \times 9.8 \times (0.643 - 0.7 \times 0.766) + 80 \times 9.8}{140} \] \[ = \frac{60 \times 9.8 \times (0.643 - 0.5362) + 784}{140} \] \[ = \frac{60 \times 9.8 \times 0.1068 + 784}{140} \] \[ = \frac{60 \times 1.047 + 784}{140} \] \[ = \frac{62.82 + 784}{140} \] \[ = \frac{846.82}{140} \] \[ = 6.05 \, \text{m/s}^2 \] **Explanation:** We combined the equations for both climbers (one on the slope, the other hanging) and solved for the acceleration of the system as Nana is pulled down the slope. --- ### **b) Nana's Acceleration with Increased Friction (\(\mu_k = 2.6\))** #### **Step 1: Substitute New Values** Use the same equation as above, but \(\mu_k = 2.6\): \[ a = \frac{60 \times 9.8 \times 0.643 - 2.6 \times 60 \times 9.8 \times 0.766 + 784}{140} \] #### **Step 2: Compute Each Term** - \(2.6 \times 0.766 = 1.9916\) - \(0.643 - 1.9916 = -1.3486\) \[ 60 \times 9.8 \times (-1.3486) = 60 \times -13.217 = -793.02 \] So the numerator: \[ -793.02 + 784 = -9.02 \] Now, divide by 140: \[ a = \frac{-9.02}{140} = -0.064 \, \text{m/s}^2 \] **Explanation:** With the increased friction, the frictional force is now greater than the net force pulling Nana down the slope, so her acceleration is slightly up the slope (negative value means she is slowing down and will stop sliding). --- ### **c) How far does she slide after using the axe?** #### **Step 1: Use Kinematics** She has an initial speed \(v_0 = 0.5\) m/s **down the slope**, and a (negative) acceleration \(a = -0.064\) m/s² (up the slope), and comes to rest (\(v_f = 0\)). Use \[ v_f^2 = v_0^2 + 2 a d \] \[ 0 = (0.5)^2 + 2(-0.064)d \] \[ 0 = 0.25 - 0.128 d \] \[ 0.128 d = 0.25 \] \[ d = \frac{0.25}{0.128} = 1.95 \, \text{m} \] **Explanation:** We used the kinematic equation for constant acceleration to find the distance Nana slides before coming to rest. --- ## **Final Answers** **a) Nana’s acceleration as she is pulled:** \[ \boxed{6.05 \, \text{m/s}^2} \] **b) Nana’s acceleration after using her axe:** \[ \boxed{-0.064 \, \text{m/s}^2} \] **c) Distance Nana slides after using the axe:** \[ \boxed{1.95 \, \text{m}} \]

The problem primarily involves applying Newton's Second Law to a system of connected bodies on an inclined plane. In part (a), we analyze the forces acting on Nana and Popo, considering the components of gravity along the slope, kinetic friction opposing motion, and tension in the rope connecting them. By setting up equations for both Nana (on the slope) and Popo (hanging), and solving simultaneously, we determine Nana’s acceleration as the rope becomes taut. In part (b), we recognize that increasing Nana’s coefficient of kinetic friction alters the frictional force opposing her motion. Repeating the force balance with the new friction coefficient, we find that her acceleration becomes negative, indicating she slows down or moves uphill. Finally, in part (c), we use kinematic equations to relate Nana’s initial speed, the calculated acceleration, and the distance traveled until she comes to rest. This approach combines Newtonian dynamics with kinematic relations to analyze motion on an inclined surface with varying friction.