Question

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 _________ $N m^{-1}$. ($Use g=10 m s^{-2}$)

Answer 1

Correct Answer: 10

To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement it causes: \( F = kx \), where \( F \) is the force, \( k \) is the force constant, and \( x \) is the displacement.
The force acting on the sphere at equilibrium is due to gravity, given by \( F = mg \), where \( m = 0.5 \) kg (convert 500 g to kg) and \( g = 10 \, \text{m/s}^2 \). Therefore, \( F = 0.5 \times 10 = 5 \, \text{N} \).
The displacement \( x = 0.5 \, \text{m} \) (convert 50 cm to meters).
Substituting these values into Hooke's Law gives \( 5 = k \times 0.5 \).
Solving for \( k \), we get \( k = \frac{5}{0.5} = 10 \, \text{N/m} \).
Therefore, the force constant of the spring is \( 10 \, \text{N/m} \). This value falls within the expected range of 10 to 10, confirming our calculation is correct.