The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A body starting from the rest at top comes back to rest at the bottom if the coefficient of friction for the lower half is given

The correct answer is μ = 2tan θ.

Let the total length of the inclined plane be 2L, so the upper half (smooth) has length L and the lower half (rough) also has length L. The plane is inclined at angle θ to the horizontal. A body starts from rest at the top and comes back to rest at the bottom.

Step 1: Analyze the downward journey

The body starts from rest at the top and slides down the entire length 2L.

On the upper half (smooth, length L):
Only the component of gravity acts along the plane.
Net acceleration = g\sin \theta

Using v^2=u^2+2as for the upper half (u = 0):

v^2=0+2\left(g\sin \theta \right)L=2gL\sin \theta

This is the velocity at the midpoint (entering the rough lower half).

On the lower half (rough, length L):
Net deceleration = g\sin \theta +\mu g\cos \theta (friction acts upward, opposing motion which is downward)

Let the velocity at the bottom after traveling the lower half be v_b. Using v^2=u^2-2as:

v_b^2=2gL\sin \theta -2(g\sin \theta +\mu g\cos \theta )L

v_b^2=2gL\sin \theta -2gL\sin \theta -2\mu gL\cos \theta

v_b^2=-2\mu gL\cos \theta

Wait — this gives a negative value, which means the body cannot reach the bottom if we assume that. Let me reconsider. The friction is decelerating while going down, but gravity helps. So friction must be less than gravity component for the body to reach the bottom. The body does reach the bottom with some velocity v_b:

v_b^2=2gL\sin \theta -2gL(\sin \theta +\mu \cos \theta )

v_b^2=-2\mu gL\cos \theta ... (this cannot be right for a positive v²)

This means the condition "comes back to rest at the bottom" must be satisfied through energy methods. Let us use the Work-Energy Theorem for the entire journey.

Step 2: Use Energy Method for the Complete Journey (Down + Up)

The body starts from rest at the top and ends at rest at the bottom. So, net change in kinetic energy = 0.

By the Work-Energy Theorem for the entire trip:

\text{Work done by gravity}+\text{Work done by friction}=0

Work done by gravity (entire trip, down + up):
The body travels down a height h=2L\sin \theta and then up the same height, so net vertical displacement = 0. Hence:

\text{Net work by gravity}=0

Therefore, the work done by friction must also equal zero — but friction always does negative work, so this cannot be zero unless we carefully track the journey.

Let me restate correctly. Since the body starts and ends at rest, by Work-Energy Theorem for the entire round trip:

\text{Work by gravity (down)}+\text{Work by gravity (up)}+\text{Work by friction (down, rough part)}+\text{Work by friction (up, rough part)}=0

Work by gravity going down = mg\left(2L\right)\sin \theta (positive, force and displacement in same direction)

Work by gravity going up = -mg\left(2L\right)\sin \theta (negative)

So net work by gravity = 0. ✓

Now, friction only acts on the lower half (length L) both during downward and upward travel:

Work by friction going down through rough part = -\mu mg\cos \theta ·L

Work by friction going up through rough part = -\mu mg\cos \theta ·L

(Friction always opposes motion, so it always does negative work.)

Total work by friction = -2\mu mgL\cos \theta

Setting total work = 0 only accounts for kinetic energy change. The correct energy equation for the round trip is:

\text{Loss in PE}=\text{Work done against friction}

Wait — the body goes down overall (starts at top, ends at bottom), so there IS a net loss in potential energy.

Step 3: Apply Energy Conservation Correctly

Since the body starts at rest at the top and ends at rest at the bottom:

Height descended (net) = 2L\sin \theta

Net loss in PE = mg·2L\sin \theta

This energy is completely lost to friction (since ΔKE = 0):

mg·2L\sin \theta =\text{Total work done against friction}

But friction only acts on the lower half (length L), and the body traverses this rough lower half twice (once going down, once going up):

\text{Total work done against friction}=\mu mg\cos \theta ·L+\mu mg\cos \theta ·L=2\mu mgL\cos \theta

Setting them equal:

mg·2L\sin \theta =2\mu mgL\cos \theta

Dividing both sides by 2mgL\cos \theta:

\frac{\sin \theta }{\cos \theta }=\mu

\mu =\tan \theta

Hmm — this gives μ = tan θ, not 2tan θ. Let me re-examine the problem setup. The issue is in the upward journey — the body goes back to the bottom of the incline, meaning it only travels a total distance of L upward (the lower rough half), NOT the full 2L back up. Let me re-read: "comes back to rest at the bottom."

Step 4: Correct Interpretation of the Journey

The body starts at the top, slides all the way to the bottom of the incline (traveling 2L), and then bounces back up (due to some mechanism? No — the problem means the body slides down and then, because of the incline's geometry, comes to rest at the bottom on the way down).

Actually, re-reading: the body starts from rest at top and "comes back to rest at the bottom." This means the body slides down the full length and comes to rest exactly at the bottom. There is NO return journey.

Step 5: Energy Method for Downward Journey Only

Body slides from top to bottom (total length 2L). It starts at rest and ends at rest.

Net loss in KE = 0, so:

\text{Work by gravity}=\text{Work done against friction}

mg·2L\sin \theta =\mu mg\cos \theta ·L

(Friction only acts over the lower half of length L)

Dividing both sides by mgL:

2\sin \theta =\mu \cos \theta

Dividing both sides by \cos \theta:

\mu =\frac{2\sin \theta }{\cos \theta }

\mu =2\tan \theta

Step 6: Physical Verification

Let's verify this makes physical sense. On the upper smooth half, the body accelerates freely. On the lower rough half, friction decelerates the body. For the body to come to rest exactly at the bottom:

• KE gained on upper half = mgL\sin \theta
• On lower half, net force = mg\sin \theta -\mu mg\cos \theta (deceleration if μ > tan θ, acceleration if μ < tan θ)
• KE gained on lower half = mgL\sin \theta -\mu mgL\cos \theta
• Total KE at bottom = 2mgL\sin \theta -\mu mgL\cos \theta =0
• ⇒ \mu =\frac{2\sin \theta }{\cos \theta }=2\tan \theta ✓

This confirms that the coefficient of friction for the lower rough half must be μ = 2tan θ for the body to come exactly to rest at the bottom after starting from rest at the top. The factor of 2 arises because the body gains kinetic energy over the full length 2L due to gravity, but friction only acts over half the length L, so it must be twice as effective per unit length to dissipate all the energy.