Solution to Question 27
1. Condition for Sliding
The block starts sliding when the angle of inclination $\theta$ reaches the “angle of repose”. At this instant, the component of gravity down the plane equals the maximum static friction. $$ mg \sin\theta = \mu_s mg \cos\theta $$ $$ \tan\theta = \mu_s $$ From this, we can derive the sine and cosine values using a right-angled triangle with opposite $\mu_s$ and adjacent $1$: $$ \sin\theta = \frac{\mu_s}{\sqrt{1 + \mu_s^2}}, \quad \cos\theta = \frac{1}{\sqrt{1 + \mu_s^2}} $$
The block starts sliding when the angle of inclination $\theta$ reaches the “angle of repose”. At this instant, the component of gravity down the plane equals the maximum static friction. $$ mg \sin\theta = \mu_s mg \cos\theta $$ $$ \tan\theta = \mu_s $$ From this, we can derive the sine and cosine values using a right-angled triangle with opposite $\mu_s$ and adjacent $1$: $$ \sin\theta = \frac{\mu_s}{\sqrt{1 + \mu_s^2}}, \quad \cos\theta = \frac{1}{\sqrt{1 + \mu_s^2}} $$
2. Motion Dynamics
Once the block starts moving, the friction changes from static ($\mu_s$) to kinetic ($\mu_k$). Since $\mu_k < \mu_s$, there is a net downward force. The net acceleration $a$ down the plane is: $$ a = g \sin\theta - \mu_k g \cos\theta $$ Substituting the values of $\sin\theta$ and $\cos\theta$: $$ a = g \left( \frac{\mu_s}{\sqrt{1 + \mu_s^2}} \right) - \mu_k g \left( \frac{1}{\sqrt{1 + \mu_s^2}} \right) $$ $$ a = \frac{g (\mu_s - \mu_k)}{\sqrt{1 + \mu_s^2}} $$
Once the block starts moving, the friction changes from static ($\mu_s$) to kinetic ($\mu_k$). Since $\mu_k < \mu_s$, there is a net downward force. The net acceleration $a$ down the plane is: $$ a = g \sin\theta - \mu_k g \cos\theta $$ Substituting the values of $\sin\theta$ and $\cos\theta$: $$ a = g \left( \frac{\mu_s}{\sqrt{1 + \mu_s^2}} \right) - \mu_k g \left( \frac{1}{\sqrt{1 + \mu_s^2}} \right) $$ $$ a = \frac{g (\mu_s - \mu_k)}{\sqrt{1 + \mu_s^2}} $$
3. Kinematics
We need to find the speed $v$ after sliding a distance $l$, starting from rest ($u=0$). Using the equation of motion $v^2 = u^2 + 2al$: $$ v^2 = 0 + 2 \left( \frac{g (\mu_s – \mu_k)}{\sqrt{1 + \mu_s^2}} \right) l $$ $$ v = \sqrt{\frac{2 (\mu_s – \mu_k) g l}{\sqrt{1 + \mu_s^2}}} $$
We need to find the speed $v$ after sliding a distance $l$, starting from rest ($u=0$). Using the equation of motion $v^2 = u^2 + 2al$: $$ v^2 = 0 + 2 \left( \frac{g (\mu_s – \mu_k)}{\sqrt{1 + \mu_s^2}} \right) l $$ $$ v = \sqrt{\frac{2 (\mu_s – \mu_k) g l}{\sqrt{1 + \mu_s^2}}} $$
Answer:
The speed acquired by the block is $ \displaystyle \sqrt{\frac{2(\mu_s – \mu_k)gl}{\sqrt{1+\mu_s^2}}} $.
The speed acquired by the block is $ \displaystyle \sqrt{\frac{2(\mu_s – \mu_k)gl}{\sqrt{1+\mu_s^2}}} $.