A block of mass m lying on a smooth horizontal surface is attached to a spring (of negligible mass)

Q: Why is the speed maximum when the net force is zero?

Acceleration is the rate of change of velocity. When net force is positive, acceleration is positive, so velocity increases. When net force becomes zero, acceleration is zero, meaning velocity stops increasing. Right after this point, the spring force becomes greater than the pulling force, causing negative acceleration (deceleration), which decreases the velocity. Hence, the peak velocity occurs exactly where the net force is zero.

Q: Why can't we just use $F = ma$ and the standard equations of motion like $v^2 = u^2 + 2as$?

The standard equations of motion ($v = u + at$, etc.) are strictly valid only when acceleration is constant. In this problem, the spring force ($kx$) changes as the block moves, meaning the net force ($F - kx$) and therefore the acceleration are variable. You must use the Work-Energy Theorem or Calculus to handle variable forces.

Q: Does the block stop at $x = F/k$?

No! At $x = F/k$, the block has its maximum kinetic energy and maximum speed. Because it has inertia, it will overshoot this equilibrium position and continue moving to the right until its kinetic energy is completely converted into the spring's potential energy.

Q: What is the maximum extension of the spring before the block momentarily stops?

You can find this by setting the final kinetic energy to zero in the Work-Energy equation: $F \cdot x_{max} - \frac{1}{2}kx_{max}^2 = 0$. Solving this gives $x_{max} = 2F/k$. The block stops momentarily here and then gets pulled back by the spring.

Q: How does the Simple Harmonic Motion (SHM) approach work for this problem?

Applying a constant force to a spring-mass system doesn't change its angular frequency ($\omega = \sqrt{k/m}$); it only shifts the equilibrium (mean) position by $\Delta x = F/k$. Since the block is released from rest at the natural length (which is now at a distance $F/k$ from the new mean position), the amplitude of oscillation is $A = F/k$. The maximum speed in SHM is $v_{max} = A\omega$, which gives the same result!

Question

A block of mass $m$ lying on a smooth horizontal surface is attached to a spring (of negligible mass) of spring constant $k$. The other end of the spring is fixed as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force $F$, the maximum speed of the block is

Options:

(A) $\frac{\pi F}{\sqrt{mk}}$

(B) $\frac{F}{\sqrt{mk}}$

(C) $\frac{2F}{\sqrt{mk}}$

(D) $\frac{F}{\pi\sqrt{mk}}$

Sigma Percentile · Accepted Answer

### The Physical Setup

Imagine a block resting peacefully on a smooth, frictionless table. It is attached to a relaxed spring. Suddenly, a constant force $$F$$ starts pulling the block to the right.

What happens next? The block begins to accelerate because of the pull. However, as it moves forward, the spring stretches and starts pulling back with a restoring force $$kx$$.

This creates a dynamic tug-of-war. The block will keep speeding up as long as the pulling force is stronger than the spring's resistance.

### The Master Equation: Work-Energy Theorem

To find the speed of the block at any position $$x$$, the **Work-Energy Theorem** is our most powerful tool. It states that the net work done by all forces equals the change in kinetic energy.

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

Here, the forces doing work are the constant pull $$F$$ and the restoring spring force. The constant force $$F$$ pulls the block through a distance $$x$$, doing positive work $$F \cdot x$$.

The spring, however, opposes this motion. The work done by the spring is negative and equals $$-\frac{1}{2}kx^2$$. This total work gives the block its kinetic energy.

$$F \cdot x - \frac{1}{2}kx^2 = \frac{1}{2}mv^2$$

### Finding the Maximum Speed

Now, we need to find the condition for maximum speed. The speed increases as long as the pulling force is greater than the spring force.

The moment the spring force equals the pulling force, the net force becomes zero. This means the acceleration is zero, and the speed hits its absolute maximum!

$$F_{	ext{net}} = 0 \implies F - kx = 0$$

Solving this gives us the critical position where the speed is maximum:

$$x = \frac{F}{k}$$

### The Final Calculation

Let's substitute this critical position back into our work-energy equation to find the maximum kinetic energy.

$$F\left(\frac{F}{k}ight) - \frac{1}{2}k\left(\frac{F}{k}ight)^2 = \frac{1}{2}mv_{	ext{max}}^2$$

Simplifying the left side, we get:

$$\frac{F^2}{k} - \frac{F^2}{2k} = \frac{1}{2}mv_{	ext{max}}^2$$

This beautifully simplifies to just $$\frac{F^2}{2k}$$. Notice how the $$\frac{1}{2}$$ factor cancels out on both sides!

$$\frac{F^2}{2k} = \frac{1}{2}mv_{	ext{max}}^2$$

Finally, we isolate $$v_{	ext{max}}$$. Taking the square root of both sides, we find our elegant final answer:

$$v_{	ext{max}} = \frac{F}{\sqrt{mk}}$$

### An Elegant Alternative: Simple Harmonic Motion

You could also solve this using the concepts of Simple Harmonic Motion (SHM). The constant force simply shifts the mean position of the spring-mass system by a distance $$\frac{F}{k}$$.

Since the block started from rest at the natural length, the amplitude $$A$$ of the oscillation is exactly this shift, $$A = \frac{F}{k}$$.

The maximum velocity in SHM is simply the amplitude multiplied by the angular frequency $$\omega$$.

$$v_{	ext{max}} = A\omega = \left(\frac{F}{k}ight)\sqrt{\frac{k}{m}} = \frac{F}{\sqrt{mk}}$$

Both paths lead to the exact same beautiful result!

Select Answer:

Visualized Solution

\text{Analyzing the Physical System}

\text{Work-Energy Theorem}

\text{Work Done by Forces}

\text{Condition for } v_{\text{max}}

\text{Calculating Maximum Kinetic Energy}

\text{Simplifying the Expression}

\text{Final Maximum Speed}

\text{Alternative Approach: SHM}

The Sigma Insight: Work Done by Forces

The Physical Setup

The Master Equation

Finding the Maximum Speed

The Final Calculation

An Elegant Alternative

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