A 0.5 kg block slides from the point A (see fig.) on a horizontal track with an initial speed of 3 m

Q: Wait, why don't we consider any energy loss from $A$ to $B$?

The problem explicitly states that the part $AB$ of the track is frictionless. Without friction or any other non-conservative forces acting in the horizontal direction, the block's kinetic energy remains perfectly conserved until it reaches point $B$.

Q: Why is the distance for friction taken as $BD + x$ and not just $BD$?

Friction acts on the block as long as it is moving on the rough surface. The block doesn't stop exactly at $D$; it continues to move forward, compressing the spring by a distance $x$. Therefore, it slides on the rough surface for a total distance of $BD + x$.

Q: Why do we use kinetic friction during the motion but check static friction at the end?

While the block is sliding, the surfaces are in relative motion, so kinetic friction ($\mu_k$) applies. However, once the block comes to a momentary halt at maximum compression, it is no longer moving. To see if it starts moving again (rebounds), we must compare the pushing force of the spring against the maximum static friction ($\mu_s$), which resists the *start* of motion.

Q: What would happen if the spring force was greater than the maximum static friction?

If $kx > \mu_s mg$, the spring would overcome the static friction and push the block back to the left. The block would then slide backwards, and we would have to set up a new Work-Energy equation to find where it finally comes to rest after rebounding.

Q: Why is the work done by the spring negative?

Work is the dot product of force and displacement ($W = \vec{F} \cdot \vec{d}$). As the block moves to the right (compressing the spring), the spring exerts a restoring force to the left. Since the force and displacement are in opposite directions (angle is $180^\circ$), the work done is negative.

Question

A $0.5	ext{ kg}$ block slides from the point $A$ (see fig.) on a horizontal track with an initial speed of $3	ext{ m/s}$ towards a weightless horizontal spring of length $1	ext{ m}$ and force constant $2	ext{ N/m}$. The part $AB$ of the track is frictionless and the part $BC$ has the coefficients of static and kinetic friction as $0.22$ and $0.2$ respectively. If the distances $AB$ and $BD$ are $2	ext{ m}$ and $2.14	ext{ m}$ respectively, find the total distance through which the block moves before it comes to rest completely. (Take $g = 10	ext{ m/s}^2$).

Sigma Percentile · Accepted Answer

Imagine a block sliding freely on a perfectly smooth surface. It has kinetic energy, momentum, and a clear path ahead. But then, it hits a rough patch. Friction starts gnawing away at its energy. As if that wasn't enough, it crashes into a spring, which pushes back with increasing force until the block finally surrenders and comes to a halt. This classic physics problem is a beautiful interplay of kinetic energy, non-conservative forces, and elastic potential energy.

### Analyzing the Setup

Our block of mass $m = 0.5	ext{ kg}$ starts at point $A$ with an initial velocity $v = 3	ext{ m/s}$. The journey from $A$ to $B$ is frictionless, meaning the block cruises to $B$ without losing a single joule of energy.

However, the segment from $B$ onwards is rough, with a coefficient of kinetic friction $\mu_k = 0.2$. The block travels a distance $BD = 2.14	ext{ m}$ before it even touches the spring. Once it hits the spring at $D$, it compresses it by an unknown distance $x$ before coming to a momentary stop. During this entire rough journey (distance $BD + x$), kinetic friction is doing negative work. Simultaneously, the spring is also doing negative work as it gets compressed.

### The Master Equation

To solve this, we deploy the ultimate weapon of mechanics: the **Work-Energy Theorem**. It states that the net work done on an object equals its change in kinetic energy.

$$W_{	ext{net}} = \Delta K$$

Here, the forces doing work are kinetic friction and the spring force. Both oppose the motion, so their work is negative. The initial kinetic energy is $\frac{1}{2}mv^2$, and the final kinetic energy is zero.

$$- \mu_k mg (BD + x) - \frac{1}{2}kx^2 = 0 - \frac{1}{2}mv^2$$

### The Calculation

Let's plug in our known values. We have $m = 0.5	ext{ kg}$, $v = 3	ext{ m/s}$, $\mu_k = 0.2$, $g = 10	ext{ m/s}^2$, $BD = 2.14	ext{ m}$, and $k = 2	ext{ N/m}$.

$$- (0.2)(0.5)(10)(2.14 + x) - \frac{1}{2}(2)x^2 = - \frac{1}{2}(0.5)(3)^2$$

Simplifying the terms, the friction force $\mu_k mg$ is exactly $1	ext{ N}$. The initial kinetic energy is $2.25	ext{ J}$.

$$1 \cdot (2.14 + x) + x^2 = 2.25$$

Rearranging this into a standard quadratic equation form:

$$x^2 + x - 0.11 = 0$$

Using the quadratic formula to find the roots:

$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-0.11)}}{2} = \frac{-1 \pm \sqrt{1.44}}{2}$$

$$x = \frac{-1 \pm 1.2}{2}$$

Since compression must be a positive distance, we discard the negative root. Thus, $x = 0.1	ext{ m}$. The spring compresses by exactly $10	ext{ cm}$.

### The Rebound Check

Here is where many students fall into a trap. The block has stopped, but will it stay there? The compressed spring is pushing back with a restoring force $F_s = kx$. If this force is stronger than the maximum static friction $f_{s,	ext{max}} = \mu_s mg$, the block will bounce back!

Let's calculate both forces. The restoring force of the spring is:

$$F_s = 2 \cdot 0.1 = 0.2	ext{ N}$$

The maximum static friction (using $\mu_s = 0.22$) is:

$$f_{s,	ext{max}} = 0.22 \cdot 0.5 \cdot 10 = 1.1	ext{ N}$$

Since $0.2	ext{ N} < 1.1	ext{ N}$, the static friction is more than capable of holding the block in place. The block does not rebound; it is permanently stuck.

### Final Calculation

The total distance covered by the block from its starting point $A$ to its final resting place is the sum of the frictionless part $AB$, the rough part before the spring $BD$, and the compression $x$.

$$d = AB + BD + x = 2 + 2.14 + 0.1 = 4.24	ext{ m}$$

And there we have it! A perfect $4.24	ext{ m}$ journey, mapped out by the elegant laws of energy conservation and friction.

Enter Numerical Value:

Visualized Solution

Initial Setup and Energy Considerations

Work-Energy Theorem

Formulating the Equation

Substituting Values

Simplifying the Equation

Solving for Compression $x$

Checking for Rebound

Calculating Forces at Rest

Final Distance Calculation

The Sigma Insight: Work Done by Forces

Analyzing the Setup

The Master Equation

The Calculation

The Rebound Check

Final Calculation

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