Let's **break down the problem** and solve it step by step using both requested methods. --- ### **Given:** - Mass of object: \( m \) - Slope height: \( h \) - Slope horizontal length: \( L \) - Coefficient of kinetic friction: \( k \) (often denoted \( \mu_k \)) - Find: **Speed at bottom** of the slope (\( v \)), in terms of \( m, h, L, k \). --- ## **Step 1: Find the length and angle of the slope** - Let \( \theta \) be the angle of the slope. The length of the slope (hypotenuse): \[ d = \frac{h}{\sin\theta} \] But also, \(\cos\theta = \frac{L}{d}\) and \(\sin\theta = \frac{h}{d}\). So, \[ d = \sqrt{L^2 + h^2} \] and \[ \sin\theta = \frac{h}{\sqrt{L^2 + h^2}} \] --- ## **(a) Using Dynamics and Kinematics** **Step 2: Forces** - Gravity down the slope: \( mg\sin\theta \) - Friction up the slope: \( f_k = kN = k mg \cos\theta \) **Net force down the slope:** \[ F_{net} = mg\sin\theta - k mg \cos\theta \] **Step 3: Acceleration** \[ a = \frac{F_{net}}{m} = g(\sin\theta - k\cos\theta) \] --- **Step 4: Kinematics** Initial velocity \( u = 0 \), distance along slope \( s = d = \sqrt{L^2 + h^2} \): \[ v^2 = u^2 + 2as \implies v^2 = 2 a d \] Plug in \( a \) and \( d \): \[ v^2 = 2g(\sin\theta - k\cos\theta)\, \sqrt{L^2 + h^2} \] Substitute \( \sin\theta = \frac{h}{\sqrt{L^2 + h^2}} \), \( \cos\theta = \frac{L}{\sqrt{L^2 + h^2}} \): \[ v^2 = 2g\left( \frac{h - kL}{\sqrt{L^2 + h^2}} \right)\sqrt{L^2 + h^2} \] \[ v^2 = 2g(h - kL) \] \[ \boxed{v = \sqrt{2g(h - kL)}} \] --- ## **(b) Using Energy** **Step 1: Initial Energy** - Potential energy at top: \( U_i = mgh \) **Step 2: Work done by friction** - Frictional force: \( f_k = k mg \cos\theta \) - Distance moved: \( d = \sqrt{L^2 + h^2} \) - Work by friction: \( W_f = -f_k d = -kmg\cos\theta \cdot d \) \[ W_f = -kmg\left(\frac{L}{\sqrt{L^2 + h^2}}\right)\sqrt{L^2 + h^2} = -kmgL \] **Step 3: Final Energy** - Kinetic energy at bottom: \( K_f = \frac{1}{2}mv^2 \) **Step 4: Conservation of Energy** \[ mgh - kmgL = \frac{1}{2}mv^2 \] \[ 2g(h - kL) = v^2 \] \[ \boxed{v = \sqrt{2g(h - kL)}} \] --- ### **Final Answer (Both Methods):** \[ \boxed{v = \sqrt{2g(h - kL)}} \] where: - \( v \) = speed at the bottom of the slope - \( g \) = acceleration due to gravity - \( h \) = height of the slope - \( L \) = horizontal length of the slope - \( k \) = coefficient of kinetic friction

\[ \boxed{v = \sqrt{2g(h - kL)}} \]