Solution
Step 1: Analyze Forces and Maximum Speed
The block accelerates down the plane as long as the component of gravity along the incline ($mg \sin \theta$) is greater than the friction force ($f$).
Friction is given by $f = \mu N = (kx) (mg \cos \theta)$, where $x$ is the distance slid.
The speed is maximum when the acceleration is zero, i.e., when the net force is zero: $$ mg \sin \theta = k x_0 mg \cos \theta $$ $$ \tan \theta = k x_0 \implies x_0 = \frac{\tan \theta}{k} $$ Here $x_0$ is the distance where speed is maximum.
The block accelerates down the plane as long as the component of gravity along the incline ($mg \sin \theta$) is greater than the friction force ($f$).
Friction is given by $f = \mu N = (kx) (mg \cos \theta)$, where $x$ is the distance slid.
The speed is maximum when the acceleration is zero, i.e., when the net force is zero: $$ mg \sin \theta = k x_0 mg \cos \theta $$ $$ \tan \theta = k x_0 \implies x_0 = \frac{\tan \theta}{k} $$ Here $x_0$ is the distance where speed is maximum.
Step 2: Use Work-Energy Theorem for Total Motion
The block starts from rest and stops after sliding a vertical height $h$. Let the total distance slid be $L$. From geometry, $L = h / \sin \theta$.
Apply Work-Energy Theorem from start to stop ($v=0$ to $v=0$): $$ W_{gravity} + W_{friction} = 0 $$ $$ mg(L \sin \theta) – \int_0^L (kx mg \cos \theta) dx = 0 $$ $$ mg L \sin \theta – mg \cos \theta \left[ \frac{kx^2}{2} \right]_0^L = 0 $$ $$ L \sin \theta = \frac{1}{2} k L^2 \cos \theta $$ $$ \sin \theta = \frac{1}{2} k L \cos \theta \implies \frac{\tan \theta}{k} = \frac{L}{2} $$
The block starts from rest and stops after sliding a vertical height $h$. Let the total distance slid be $L$. From geometry, $L = h / \sin \theta$.
Apply Work-Energy Theorem from start to stop ($v=0$ to $v=0$): $$ W_{gravity} + W_{friction} = 0 $$ $$ mg(L \sin \theta) – \int_0^L (kx mg \cos \theta) dx = 0 $$ $$ mg L \sin \theta – mg \cos \theta \left[ \frac{kx^2}{2} \right]_0^L = 0 $$ $$ L \sin \theta = \frac{1}{2} k L^2 \cos \theta $$ $$ \sin \theta = \frac{1}{2} k L \cos \theta \implies \frac{\tan \theta}{k} = \frac{L}{2} $$
Step 3: Connect Maximum Speed Position to Total Distance
Comparing the result from Step 1 ($x_0 = \tan \theta / k$) and Step 2 ($L/2 = \tan \theta / k$), we find: $$ x_0 = \frac{L}{2} $$ The maximum speed occurs exactly halfway down the path.
Comparing the result from Step 1 ($x_0 = \tan \theta / k$) and Step 2 ($L/2 = \tan \theta / k$), we find: $$ x_0 = \frac{L}{2} $$ The maximum speed occurs exactly halfway down the path.
Step 4: Calculate Maximum Speed
Apply Work-Energy Theorem from the start ($x=0$) to the point of maximum speed ($x=x_0$): $$ \frac{1}{2}mv_{max}^2 = W_{gravity} – W_{friction} $$ $$ \frac{1}{2}mv_{max}^2 = mg(x_0 \sin \theta) – \int_0^{x_0} kx mg \cos \theta dx $$ $$ \frac{1}{2}mv_{max}^2 = mg x_0 \sin \theta – \frac{1}{2} k mg \cos \theta x_0^2 $$ Substitute $k x_0 \cos \theta = \sin \theta$ (from Step 1 equilibrium condition) into the friction term: $$ \frac{1}{2}mv_{max}^2 = mg x_0 \sin \theta – \frac{1}{2} mg x_0 (\sin \theta) $$ $$ \frac{1}{2}mv_{max}^2 = \frac{1}{2} mg x_0 \sin \theta $$ $$ v_{max}^2 = g x_0 \sin \theta $$ Since $x_0 = L/2$ and $L \sin \theta = h$, the vertical drop at $x_0$ is $h/2$: $$ x_0 \sin \theta = \frac{L}{2} \sin \theta = \frac{h}{2} $$ $$ v_{max} = \sqrt{\frac{gh}{2}} $$
Apply Work-Energy Theorem from the start ($x=0$) to the point of maximum speed ($x=x_0$): $$ \frac{1}{2}mv_{max}^2 = W_{gravity} – W_{friction} $$ $$ \frac{1}{2}mv_{max}^2 = mg(x_0 \sin \theta) – \int_0^{x_0} kx mg \cos \theta dx $$ $$ \frac{1}{2}mv_{max}^2 = mg x_0 \sin \theta – \frac{1}{2} k mg \cos \theta x_0^2 $$ Substitute $k x_0 \cos \theta = \sin \theta$ (from Step 1 equilibrium condition) into the friction term: $$ \frac{1}{2}mv_{max}^2 = mg x_0 \sin \theta – \frac{1}{2} mg x_0 (\sin \theta) $$ $$ \frac{1}{2}mv_{max}^2 = \frac{1}{2} mg x_0 \sin \theta $$ $$ v_{max}^2 = g x_0 \sin \theta $$ Since $x_0 = L/2$ and $L \sin \theta = h$, the vertical drop at $x_0$ is $h/2$: $$ x_0 \sin \theta = \frac{L}{2} \sin \theta = \frac{h}{2} $$ $$ v_{max} = \sqrt{\frac{gh}{2}} $$
Conclusion:
The maximum speed of the block is $\sqrt{\frac{gh}{2}}$.
The maximum speed of the block is $\sqrt{\frac{gh}{2}}$.