Q.47 A massless ideal spring is hanging vertically. A sphere of mass of 500 g, suspended
from the spring, stretches the spring from its initial position by 50 cm when it
reaches equilibrium. The force constant of the spring is _______ N m–1. (Use g=10
m s–2)
The force constant of the spring is 10 N/m. At equilibrium, the gravitational force balances the spring force.
Problem Analysis
A 500 g (0.5 kg) sphere stretches the massless ideal spring by 50 cm (0.5 m) vertically. Using g = 10 m/s², the weight mg equals the spring force kx.
Step-by-Step Solution
Convert mass to kg: m = 0.5 kg.
Extension x = 0.5 m, mg = 0.5 × 10 = 5 N.
Hooke’s law at equilibrium: k = mg / x = 5 / 0.5 = 10 N/m.
Key Equation
Spring force F_s = kx equals mg:
k=mgx
Verification
No oscillation occurs; equilibrium holds as net force is zero. Massless spring ignores its weight.
Introduction to Force Constant in Hanging Spring
In physics problems like a massless ideal spring hanging vertically with a 500g sphere causing 50cm stretch at equilibrium, the force constant (k) measures spring stiffness using Hooke’s law. This calculation applies g=10 m/s² and balances weight against restoring force, vital for CSIR NET Life Sciences/Physics crossovers in biomechanics.
Core Concept: Equilibrium in Vertical Spring
At equilibrium, upward spring force kx = downward weight mg.
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No net force, no acceleration.
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Extension x = 50 cm = 0.50 m from natural length.
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Mass m = 500 g = 0.50 kg.
Detailed Calculation: Force Constant Value
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Compute weight: mg = 0.50 kg × 10 m/s² = 5 N.
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Apply Hooke’s law: k = mg / x = 5 N / 0.50 m = 10 N/m.
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Units confirm: N/m (or N⋅m⁻¹).
| Parameter | Value | Formula Used |
|---|---|---|
| Mass (m) | 0.5 kg | 500 g converted |
| g | 10 m/s² | Given |
| Weight (mg) | 5 N | mg |
| Extension (x) | 0.5 m | 50 cm |
| Force Constant (k) | 10 N/m | mg/x |
Common Mistakes and Explanations
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Error 1: Using cm without conversion—yields k=100 N/m (wrong). Always SI units: meters.
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Error 2: Ignoring equilibrium—some confuse with oscillation amplitude (max extension 2mg/k). Here, static balance only.
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Error 3: Forgetting mass in kg—500 g as 500 kg absurdly inflates k.
No options provided, but typical MCQ distractors: 5, 100, 0.1 N/m fail unit/equilibrium checks.
Applications in CSIR NET and Beyond
Force constant hanging spring problems test Hooke’s law integration with gravity, relevant for protein folding (spring-like models) or plant cell turgor in life sciences. Practice variations: change mass to 1 kg (k=20 N/m) or g=9.8 (k=9.8 N/m).
Advanced Insights
For oscillations post-displacement, equilibrium shifts reference: effective F = -kΔx (Δx from stretched position). Time period T = 2π√(m/k) = 2π√(0.5/10) ≈ 1.41 s. Ideal assumptions: no damping, linear Hooke’s range.
