Question

A spring of force constant \(K=1000\,N/m\) is compressed by \(5\,cm\). If a mass of \(2.5\,kg\) is attached and released, find its speed as it passes the equilibrium position.

• A. \(0.5\,m/s\)
• B. \(1.0\,m/s\)
• C. \(1.5\,m/s\)
• D. \(2.0\,m/s\)

Hint:

For a spring-mass system, remember: \[ \boxed{\frac12Kx^2=\frac12mv^2} \] or directly, \[ \boxed{v=x\sqrt{\frac{K}{m}}} \] At the equilibrium position, the spring has zero potential energy and the kinetic energy of the mass is maximum.

Answer 1

The Correct Option is B

Concept: A spring-mass system performs Simple Harmonic Motion (SHM). When the spring is compressed and released, the elastic potential energy stored in the spring is completely converted into the kinetic energy of the mass as it passes through the equilibrium position. According to the law of conservation of energy, \[ \boxed{\frac{1}{2}Kx^{2}=\frac{1}{2}mv^{2}} \] where \[ K=\text{Spring constant}, \] \[ x=\text{Compression of the spring}, \] \[ m=\text{Mass attached}, \] \[ v=\text{Velocity at the equilibrium position}. \] Rearranging, \[ \boxed{v=x\sqrt{\frac{K}{m}}} \]

Step 1: Write the given data.
Given, \[ K=1000\,N/m, \] \[ x=5\,cm=0.05\,m, \] \[ m=2.5\,kg. \]

Step 2: Apply the conservation of energy.
Using, \[ \frac12 Kx^2=\frac12 mv^2, \] we get, \[ v=x\sqrt{\frac{K}{m}}. \] Substituting the given values, \[ v = 0.05 \sqrt{\frac{1000}{2.5}}. \]

Step 3: Calculate the numerical value.
First, \[ \frac{1000}{2.5}=400. \] Therefore, \[ \sqrt{400}=20. \] Hence, \[ v = 0.05\times20 = 1.0\,m/s. \] Thus, \[ \boxed{v=1.0\,m/s.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (B)}}. \]

Verification: Using energy directly, \[ \frac12(1000)(0.05)^2 = \frac12(2.5)v^2, \] \[ 1.25 = 1.25v^2, \] \[ v^2=1, \] \[ v=1\,m/s. \] Thus, the answer is verified.