NLM O35

Pathfinder Physics Solution – MCQ 35

Pathfinder Physics Solution — Laws of Motion (MCQ 35)

Problem Statement:
In the setup shown, a block is placed on a frictionless floor, the cords and pulleys are ideal and each spring has stiffness k. The block is pulled away from the wall. How far will the block shift, while the pulling force is increased gradually from zero to a value F?

Corrected Technical Diagram (Perfect Geometry Alignment)

x₁ x₁ 2x₁ 2x₁ Block x F

Method 1: Kinematic Constraint & Spring Elongation (Alternative Solution)

We can systematically solve this by establishing the internal relationship between individual spring extensions and the absolute rightward displacement x of the block.

1. Variable Definitions Based on System Topology

  • Let the elongation of the top-left spring be x₁. Because this string is anchored directly to the wall, the center axle of the upper pulley shifts rightwards by exactly x₁.
  • Let the elongation of the middle-right spring connected to the upper pulley axle be x₂.
  • From the ideal force equilibrium conditions of the pulleys, the tensions dictate the remaining spring responses: the lower-left spring scales to an elongation of 2x₁, and the lower-right spring attached to the block scales to x₁.

2. Formulating the Constraint Geometry Equations

Tracking the net horizontal translation of the block through the length constraints of the upper and lower loop combinations gives us the boundary conditions:

Upper Tracking Equation: 2x₂ – 4x₁ + x = 2x₁ ⇒ x = 6x₁ – 2x₂
Lower Compatibility Equation: x₂ + 2x₁ = 6x₁ – 2x₂

Simplifying the compatibility condition directly isolates the ratio between the spring expansions:

3x₂ = 4x₁ ⇒ x₂ = (4/3)x₁

3. Shifting Calculation of the Block

Expressing the total net shift x of the block explicitly using our base translation component x₁:

x = x₂ + 2x₁ = (4/3)x₁ + 2x₁ = (10/3)x₁

The external load $F$ dynamically pulls against a cumulative resisting hookean network equivalent to:

F = 3k · x₁ ⇒ x₁ = F / (3k)

Substituting $x₁$ into our displacement relation delivers the geometric constraint solution:

x = (10/3) · (F / 3k) = 10F / (9k)

Method 2: Work-Energy Equivalency

1. Total Network Potential Energy ($U$)

Summing the elastic strain energy stored concurrently across all four identical mechanical springs:

U = ½k(x₁)² + ½k(2x₁)² + ½k(2x₁)² + ½k(x₁)²
U = ½k [ x₁² + 4x₁² + 4x₁² + x₁² ] = 5k x₁²

Substituting the load state parameters ($x₁ = \frac{F}{3k}$):

U = 5k · (F / 3k)² = 5F² / (9k)

2. Work Done Integration

Because the external pulling force is introduced gradually starting from zero up to steady-state value $F$, the external energy work profile evaluates to:

W = ½ · F · x

Equating work input directly to total internal potential storage ($W = U$):

½ · F · x = 5F² / (9k)
x = 10F / (9k)

Correct Option: (d) 10F / 9k