In the figures (a) and (b) AC, DG and GF are fixed inclined planes, BC = EF = x and AB = DE = y. A s

Q: Wait, why is the work done by gravity the same for both paths? Doesn't the block travel a longer distance on the broken path?

Gravity is a conservative force. This means the work it does depends strictly on the initial and final vertical positions, completely ignoring the actual path taken. Since the block drops by the exact same vertical height $y$ in both scenarios, gravity does the exact same amount of work, $Mgy$.

Q: Why doesn't the normal force do any work?

Work is defined as the dot product of force and displacement ($W = \vec{F} \cdot \vec{d}$). The normal force is always perpendicular ($90^\circ$) to the surface, and therefore perpendicular to the displacement of the block. Since $\cos(90^\circ) = 0$, the work done by the normal force is always zero.

Q: Is friction a conservative force since its work only depended on the horizontal distance $x$?

No, friction is strictly a non-conservative force. The fact that its work simplified to $-\mu Mgx$ is a special geometric consequence of sliding on flat inclined planes where the normal force is $Mg \cos\theta$. If the block were moving on a flat horizontal surface and took a winding, curved path, the work done by friction would depend on the total path length, not just the straight-line displacement.

Q: What if the path was a smooth curve instead of broken straight lines? Would the final speed still be the same?

Yes! You can think of a smooth curve as an infinite number of infinitesimally small straight inclines. If you use calculus to integrate the work done by friction along the curve, $\int -\mu Mg \cos\theta \, ds$, noting that $ds \cos\theta = dx$, the integral evaluates exactly to $-\mu Mgx$. The result holds for any shape of the incline as long as it doesn't curve back on itself.

Q: Does the mass of the block affect the final speed?

No, it doesn't. If you look at the final Work-Energy equation, $Mgy - \mu Mgx = \frac{1}{2}Mv^2$, you can see that the mass $M$ appears in every single term. It completely cancels out of the equation, leaving $v = \sqrt{2g(y - \mu x)}$. A heavy block and a light block will reach the bottom with the exact same speed.

Question

In the figures (a) and (b) $AC$, $DG$ and $GF$ are fixed inclined planes, $BC = EF = x$ and $AB = DE = y$. A small block of mass $M$ is released from the point $A$. It slides down $AC$ and reaches $C$ with a speed $v_C$. The same block is released from rest from the point $D$. It slides down $DGF$ and reaches the point $F$ with speed $v_F$. The coefficients of kinetic frictions between the block and both the surfaces $AC$ and $DGF$ are $\mu$. Calculate $v_C$ and $v_F$.

Sigma Percentile · Accepted Answer

### The Setup: Two Paths, One Destination

Imagine you are standing at the top of a hill and you have two choices to slide down. The first choice is a perfectly straight, smooth-looking (but still rough!) inclined plane. The second choice is a bumpy, broken path that changes its slope midway. Both paths start at the same height and end at the exact same horizontal distance.

This is exactly the scenario presented in our problem. We have a block of mass $M$ sliding down a straight incline $AC$ in figure (a), and the same block sliding down a broken incline $DGF$ in figure (b). Both paths have a total vertical drop of $y$ and a total horizontal span of $x$. The coefficient of kinetic friction $\mu$ is the same everywhere. Our mission is to find the final speeds $v_C$ and $v_F$ at the bottom of these paths.

### The Work-Energy Theorem

Whenever a problem asks for final speeds after a particle has moved through a distance under the influence of forces, your brain should immediately scream: **Work-Energy Theorem!**

The theorem is beautifully simple:
$$W_{	ext{net}} = \Delta K = K_f - K_i$$

Since the block is released from rest in both cases, the initial kinetic energy $K_i$ is zero. The net work done on the block is the sum of the work done by gravity ($W_g$) and the work done by kinetic friction ($W_f$). The normal force is always perpendicular to the displacement, so it does zero work.

### The Gravity of the Situation

Let's tackle gravity first. Gravity is a conservative force. This means it doesn't care whether you take a straight path, a broken path, or a spiral staircase. The work done by gravity depends *only* on the initial and final vertical positions.

In both figure (a) and figure (b), the block descends by a vertical height $y$. Therefore, the work done by gravity is identical for both paths:
$$W_g = Mgy$$

### The Friction Trick: The Horizontal Equivalent

Now comes the brilliant part—the work done by friction. Let's look at the straight incline $AC$ first. Let its length be $L$ and its angle of inclination be $	heta$. The normal force on the block is $N = Mg \cos	heta$. Consequently, the kinetic friction force is $f_k = \mu Mg \cos	heta$.

The work done by friction is negative because it opposes the motion:
$$W_{f, AC} = -f_k L = -\mu Mg L \cos	heta$$

But look closely at the geometry of the triangle! The term $L \cos	heta$ is exactly the horizontal base of the triangle, which is given as $x$. So, the work done by friction simplifies beautifully to:
$$W_{f, AC} = -\mu Mgx$$

What about the broken path $DGF$? We can break it down into tiny straight segments. For any segment $i$ with length $L_i$ and angle $	heta_i$, the work done by friction is $-\mu Mg L_i \cos	heta_i$. Again, $L_i \cos	heta_i$ is just the horizontal length of that specific segment, let's call it $x_i$.

To find the total work done by friction on the broken path, we sum up the work for all segments:
$$W_{f, DGF} = \sum (-\mu Mg x_i) = -\mu Mg \sum x_i$$

And what is the sum of all these horizontal segments? It is exactly the total horizontal span $x$! Therefore, the total work done by friction on the broken path is also:
$$W_{f, DGF} = -\mu Mgx$$

### The Grand Conclusion

This is a profound realization. On any inclined plane (straight or broken) with a constant coefficient of friction, the work done by friction depends **only on the total horizontal displacement**, not on the actual length of the path or its varying slopes!

Since both the work done by gravity ($Mgy$) and the work done by friction ($-\mu Mgx$) are identical for both paths, the net work done is the same. By the Work-Energy Theorem, the final kinetic energy must be the same.

$$Mgy - \mu Mgx = \frac{1}{2}Mv^2$$

Solving for $v$, we get the final speed for both cases:
$$v_C = v_F = \sqrt{2g(y - \mu x)}$$

The broken path was just an illusion to test your conceptual clarity!

Visualized Solution

The Two Inclined Planes

Work-Energy Theorem

Work Done by Gravity

Friction on Path AC

Work of Friction on AC

Calculating $v_{C}$

Friction on Path DGF

Total Friction Work on DGF

Calculating $v_{F}$

The Sigma Insight: Work Done by Forces

The Setup

The Work-Energy Theorem

The Gravity of the Situation

The Friction Trick

The Grand Conclusion

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