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Current Question (ID: 19242)

Question:

$\text{A person pushes a box on a rough horizontal platform surface. He applies a force of } 200 \text{ N over a distance of } 15 \text{ m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to } 100 \text{ N. The total distance through which the box has been moved is } 30 \text{ m. What is the work done by the person during the total movement of the box?}$

Options:

1. $5690 \text{ J}$
2. $3280 \text{ J}$
3. $2780 \text{ J}$
4. $5250 \text{ J}$

Solution:

$\text{Hint: } W = \int F \, dx$ $F = 200 \text{ N for } 0 \leq x < 15$ $= 200 - \frac{100}{15} (x - 15)$ $W = \int F \, dx$ $= \int_0^{15} 200 \, dx + \int_{15}^{30} \left(300 - \frac{100}{15} x\right) \, dx$ $= 200 \times 15 + 300 \times 15 - 2250 - \frac{100}{15} \times \left(\frac{30^2}{2} - \frac{1}{2}\right)$ $= 3000 + 4500 - 2250$ $= 5250 \text{ J}$

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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
}