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Current Question (ID: 11241)

Question:

$\text{A block of mass 4 kg hangs from a spring of spring constant } k = 400 \text{ N/m. The block is pulled down through 15 cm below the equilibrium position and released. What is its kinetic energy when the block is 10 cm below the equilibrium position? [Ignore gravity]}$

Options:

1. $5 \text{ J}$
2. $2.5 \text{ J}$ (Correct)
3. $1 \text{ J}$
4. $1.9 \text{ J}$

Solution:

$\text{Hint: Extreme position is the position of the bob at the maximum distance from the mean position so here amplitude is 15cm.}$ $\text{Step 1: Calculate angular frequency.}$ $\text{The frequency of the block performing SHM in the spring-mass system is given as}$ $\omega = \sqrt{\frac{k}{m}}$ $\text{Here,}$ $k = 400 \text{ N/m}$ $m = 4 \text{ kg}$ $\text{So, } \omega = \sqrt{\frac{k}{m}} = 10 \text{ rad/s}$ $\text{Step 2: Calculate the kinetic energy when the block is 10 cm below the equilibrium position.}$ $\text{Kinetic energy at a distance y from the mean position is given as}$ $K = \frac{1}{2}m\omega^2(A^2 - y^2)$ $\text{Here, } A = 0.15\text{m} \text{ and } y = 0.1\text{m}.$ $\text{So,}$ $K = \frac{1}{2} \times 4 \times (10)^2[(0.15)^2 - (0.1)^2]$ $= 2.5 \text{ J}$ $\text{Therefore, the kinetic energy of the block is 2.5 J.}$ $\text{Hence, option (2) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}