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Step-by-step Solution
Step 1: Understand the Problem
A block is projected up a rough inclined plane (inclination of 30°) with an initial speed v_{0} . After moving up and then sliding back down to its original position, the speed of the block becomes \frac{v_{0}}{2} . We need to find the coefficient of kinetic friction \mu between the block and plane, given that \mu = \frac{I}{1000} and we must determine the integer value of I .
Step 2: Set Up Notations and Known Facts
Mass of the block = m (cancels out eventually).
Inclination angle = 30^\circ .
Coefficient of kinetic friction = \mu .
Acceleration due to gravity = g .
Normal force on the incline = mg \cos 30^\circ .
Frictional force f = \mu \, mg \cos 30^\circ , acting opposite to the direction of motion.
Step 3: Energy Analysis for the Upward Journey
When the block travels up the incline from the bottom to its highest point:
Initial kinetic energy = \frac{1}{2} m v_{0}^{2}.
Change in potential energy = m g \,(D \sin 30^\circ) if D is the distance traveled along the incline.
Work done by friction (negative) = -\,\mu\, m g \cos 30^\circ \times D.
Final kinetic energy at the top = 0 (the block momentarily stops).
By conservation of energy on the way up:
\frac{1}{2} m v_{0}^{2} \;-\; m g \,(D \sin 30^\circ) \;-\; \mu \, m g \cos 30^\circ \, D \;=\; 0.
Rearrange to solve for D :
\frac{1}{2} m v_{0}^{2} \;=\; m g \,D \,\bigl(\sin 30^\circ + \mu \cos 30^\circ\bigr).\\
\Rightarrow D \;=\; \frac{v_{0}^{2}}{2\,g\,\bigl(\sin 30^\circ + \mu \cos 30^\circ\bigr)}.
Step 4: Energy Analysis for the Downward Journey
Now the block slides back down the same distance D , starting from rest at the top and ending with speed \frac{v_{0}}{2} at the bottom:
Initial kinetic energy at the top = 0.
Final kinetic energy at the bottom = \tfrac{1}{2} m \bigl(\tfrac{v_{0}}{2}\bigr)^2 = \frac{1}{8} m v_{0}^{2}.
Work done by gravity (positive down the plane) = +\,m g \,D \sin 30^\circ.
Work done by friction (negative) = -\,\mu \, m g \cos 30^\circ \,D.
Equating the net work to the change in kinetic energy:
m g \,D \sin 30^\circ \;-\; \mu \, m g \cos 30^\circ \,D
\;=\; \frac{1}{8} m v_{0}^{2}.
Canceling m and grouping terms:
D \;\bigl(g \sin 30^\circ - \mu \, g \cos 30^\circ\bigr) \;=\; \frac{v_{0}^{2}}{8}.
Hence,
D \;=\; \frac{v_{0}^{2}}{ 8\,g\,\bigl(\sin 30^\circ - \mu \cos 30^\circ\bigr)}.
Step 5: Equate the Two Expressions for D
The distance D is the same going up and coming down, so equate the two results:
\frac{v_{0}^{2}}{2\,g\,(\sin 30^\circ + \mu \cos 30^\circ)}
\;=\;
\frac{v_{0}^{2}}{8\,g\,(\sin 30^\circ - \mu \cos 30^\circ)}.
Cancel the common factors v_{0}^{2} and g :
\frac{1}{2\,(\sin 30^\circ + \mu \cos 30^\circ)}
\;=\;
\frac{1}{8\,(\sin 30^\circ - \mu \cos 30^\circ)}.
Cross-multiplying gives:
8\,\bigl(\sin 30^\circ - \mu \cos 30^\circ\bigr)
\;=\;
2\,\bigl(\sin 30^\circ + \mu \cos 30^\circ\bigr).
Let \sin 30^\circ = \tfrac{1}{2} and \cos 30^\circ = \tfrac{\sqrt{3}}{2} :
8 \times \tfrac{1}{2} \;-\; 8 \mu \times \tfrac{\sqrt{3}}{2}
\;=\;
2 \times \tfrac{1}{2} \;+\; 2 \mu \times \tfrac{\sqrt{3}}{2}.
4 \;-\; 4\sqrt{3}\,\mu
\;=\;
1 \;+\; \sqrt{3}\,\mu.
Combine like terms:
4 - 1 \;=\; 4\sqrt{3}\,\mu + \sqrt{3}\,\mu
\;\;\Rightarrow\;\; 3 \;=\; 5\sqrt{3}\,\mu.
Hence,
\mu = \frac{3}{5\sqrt{3}}
\;=\; \frac{\sqrt{3}}{5}
\approx 0.3464.
Step 6: Determine I
The question states \mu = \frac{I}{1000}. Therefore,
I \;=\; 1000 \,\mu \;\approx\; 1000 \times 0.3464 \;=\; 346.4.
Rounding to the nearest integer gives I = 346.
