Question

A block is kept at the top of a rough inclined plane of length $2.8\text{ m}$ and angle of inclination $\cos^{-1}(0.6)$. If the coefficient of kinetic friction between the block and the upper half of the plane is $0.3$ and between the block and the lower half of the plane is $0.5$, then the velocity with which the block reaches the bottom of the plane is (acceleration due to gravity $=10\text{ ms}^{-2}$)

• A. $1.4\text{ ms}^{-1}$
• B. $5.6\text{ ms}^{-1}$
• C. $4.2\text{ ms}^{-1}$
• D. $2.8\text{ ms}^{-1}$

Hint:

Work-Energy Theorem alternative: $v^2 = 2gL\sin\theta - 2g\cos\theta(\mu_1\frac{L}{2} + \mu_2\frac{L}{2})$ lets you calculate final velocity across multiple friction zones in a single step.

Answer 1

The Correct Option is B

Step 1: Find $\sin\theta$ from $\cos\theta$.
We are told $\cos\theta = 0.6$. Then $\sin\theta = \sqrt{1 - 0.36} = 0.8$.

Step 2: Split the incline into two halves.
Total length is $2.8$ m, so each half is $1.4$ m. The top half has friction $\mu_{1}=0.3$ and the bottom half has $\mu_{2}=0.5$.

Step 3: Acceleration on the top half.
On a rough incline $a = g(\sin\theta - \mu\cos\theta)$. \[ a_{1} = 10(0.8 - 0.3\times 0.6) = 10(0.62) = 6.2 \text{ m s}^{-2} \]
Step 4: Speed at the middle.
Start from rest, use $v^{2} = u^{2} + 2as$ for the top half: \[ v_{mid}^{2} = 0 + 2(6.2)(1.4) = 17.36 \]
Step 5: Acceleration on the bottom half.
\[ a_{2} = 10(0.8 - 0.5\times 0.6) = 10(0.5) = 5.0 \text{ m s}^{-2} \]
Step 6: Speed at the bottom.
Now use $v^{2} = v_{mid}^{2} + 2a_{2}s$ for the bottom half: \[ v^{2} = 17.36 + 2(5.0)(1.4) = 31.36 \]so $v = \sqrt{31.36} = 5.6$ m s$^{-1}$, which is option 2.
\[ \boxed{v = 5.6 \text{ m s}^{-1}} \]