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Current Question (ID: 19254)

Question:
$\text{A block of mass } 2 \text{ kg, moving with a speed of } 4 \text{ m/s on a horizontal surface enters a rough region extending from } x = 0.5 \text{ m to } x = 1.5 \text{ m. In this region, the retarding force is given by } F = -kx \text{ with } k = 12 \text{ N/m. What will be the speed of the block as it just emerges from the rough surface?}$
Options:
  • 1. $0$
  • 2. $1.5 \text{ m/s}$
  • 3. $2.0 \text{ m/s}$
  • 4. $2.5 \text{ m/s}$
Solution:
$\text{Hint: Use the work-energy theorem.}$ $\text{Step: Find the speed of the block as it just crosses the rough surface.}$ $W_{\text{net}} = \Delta K$ $\Rightarrow \int_{0.5}^{1.5} F dx = \frac{1}{2} m (v^2 - u^2)$ $\Rightarrow -\int_{0.5}^{1.5} kx dx = \frac{1}{2} m (v^2 - u^2)$ $\Rightarrow -k \left[ \frac{x^2}{2} \right]_{0.5}^{1.5} = \frac{1}{2} m (v^2 - u^2)$ $\Rightarrow -12 \left[ \frac{1.5^2}{2} - \frac{0.5^2}{2} \right] = \frac{1}{2} \times 2 (v^2 - 4^2)$ $\Rightarrow v = 2.0 \text{ m/s}$ $\text{Hence, option (3) is the correct answer.}$

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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
}