What is the Law of Conservation of Energy?

It states that the total mechanical energy (Kinetic + Potential) of an isolated system remains constant if only conservative forces act on it.

What is the formula for elastic potential energy of a spring?

The potential energy $U$ is given by $U = \frac{1}{2}k(\Delta L)^2$, where $\Delta L$ is the compression or extension from the natural length.

How is compression calculated in this problem?

Compression is the difference between natural length (2 m) and current length. Initially, it is $2 - 1 = 1$ m. At distance $x$, it is $|2 - x|$.

What is the initial kinetic energy of the block?

The block is released from rest, so its initial kinetic energy is zero.

Does friction affect the speed in this problem?

No, the problem specifies a frictionless horizontal table, so mechanical energy is conserved.

What is the spring constant $k$ given?

The spring constant $k$ is $200$ N/m.

How is the final speed $v$ isolated from the energy equation?

By equating initial potential energy to the sum of final kinetic and potential energy, then solving for $v$.

What does the term $(2-x)$ represent?

It represents the displacement from the equilibrium position (natural length) when the block is at distance $x$ from the wall.

Why is the answer expressed with a square root?

Because kinetic energy involves $v^2$, solving for $v$ requires taking the square root of the remaining terms.

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring’s natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be | JEE Main Physics

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PhysicsMechanics

QMCQ

A block of mass $2$ kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring’s natural length is $2$ m and spring constant is $200$ N/m. The block is pushed such that the length of the spring becomes $1$ m and then released. At distance $x$ m ($x < 2$) from the wall, the speed of the block will be:

A) $10[1 – (2 – x)^2]^{\frac{1}{2}}$ m/s
B) $10[1 – (2 – x)^2]^{\frac{3}{2}}$ m/s
C) $10[1 – (2 – x)^2]$ m/s
D) None of these

✅ Correct Answer

Option A

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Solution Steps

Identify Initial Energy

The natural length $L_0 = 2$ m. The spring is compressed to length $L_i = 1$ m. Compression $\Delta x_i = L_0 – L_i = 2 – 1 = 1$ m. Initial energy is purely potential:

$$E_i = \frac{1}{2}k(\Delta x_i)^2 = \frac{1}{2}(200)(1)^2 = 100 \text{ J}$$

Determine Displacement at Distance $x$

When the block is at a distance $x$ from the wall, the compression (or extension) $\Delta x_f$ from the natural length is $|L_0 – x| = |2 – x|$.

Apply Energy Conservation

Total mechanical energy remains constant because the surface is frictionless. $E_i = KE_f + PE_f$.

$$100 = \frac{1}{2}mv^2 + \frac{1}{2}k(2-x)^2$$

Substitute Values

Substitute $m = 2$ kg and $k = 200$ N/m into the equation:

$$100 = \frac{1}{2}(2)v^2 + \frac{1}{2}(200)(2-x)^2$$

$$100 = v^2 + 100(2-x)^2$$

Solve for Speed $v$

Isolate $v^2$ and then take the square root:

$$v^2 = 100 – 100(2-x)^2 = 100[1 – (2-x)^2]$$

$$v = \sqrt{100[1 – (2-x)^2]} = 10[1 – (2-x)^2]^{\frac{1}{2}} \text{ m/s}$$

Final Result: The speed is $10[1 – (2 – x)^2]^{\frac{1}{2}}$ m/s.

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Theory

1. Work-Energy Theorem and Springs

The work-energy theorem states that the work done by all forces acting on a particle equals the change in its kinetic energy. For a spring-mass system on a frictionless surface, the only work being done is by the spring force, which is a conservative force. This allows us to define a potential energy function for the spring. As the spring expands from its compressed state, it performs positive work on the block, converting stored elastic potential energy into kinetic energy. The total mechanical energy (sum of KE and PE) remains constant throughout the motion.

2. Elastic Potential Energy

Elastic potential energy is the energy stored as a result of applying a force to deform an elastic object. For a spring following Hooke’s Law ($F = -kx$), the energy stored is given by $U = \frac{1}{2}kx^2$. Crucially, ‘x’ in this formula represents the displacement from the equilibrium position (natural length), not the distance from the point of attachment. In this problem, because the wall is the origin and the natural length is $2$ m, the displacement is the difference between $2$ m and the current position of the block.

3. Simple Harmonic Motion (SHM) Characteristics

A spring-mass system is the classic example of Simple Harmonic Motion. When released from a compressed state, the block oscillates about the equilibrium point. The speed of the block is maximum at the equilibrium position (where potential energy is zero) and zero at the extreme positions (where all energy is potential). The equation derived in the solution ($v = \omega\sqrt{A^2 – y^2}$) is a general property of SHM, where $\omega = \sqrt{k/m}$ is the angular frequency and $A$ is the amplitude of oscillation.

4. Massless Springs in Mechanics

In JEE level physics, springs are typically assumed to be ‘massless’. This idealization means we do not need to account for the kinetic energy of the spring coils themselves. All the inertia of the system is concentrated in the block ($m$), and all the elasticity is concentrated in the spring ($k$). This simplifies the energy balance to a two-term sum, making it easier to solve for the velocity of the attached block at any arbitrary point in its trajectory.

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FAQs

What is the amplitude of this oscillation?

The amplitude is the maximum displacement from equilibrium, which is $2 – 1 = 1$ m.

Where is the block’s speed maximum?

The speed is maximum at the equilibrium position, which is $x = 2$ m from the wall.

What is the maximum speed?

At $x=2$, $v = 10\sqrt{1-0} = 10$ m/s.

Why did we use $(2-x)$ instead of just $x$?

Potential energy depends on the change from natural length. Since the natural length is 2 m, the deformation is $2-x$.

Does the direction of motion matter for speed?

No, speed is a scalar and depends only on the magnitude of displacement from equilibrium.

What happens if the table had friction?

Mechanical energy would not be conserved; some energy would be lost as heat, reducing the final speed.

Is the acceleration constant?

No, acceleration depends on the spring force $F = -kx$, so it varies with position.

What is the time period of oscillation?

$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{2/200} = 2\pi/10 = \pi/5$ seconds.

What if $x > 2$?

The formula still works as $(2-x)^2$ is always positive, representing the energy in the stretched spring.

Why is Option A correct?

Because it correctly represents the square root of the energy balance equation.

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