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Current Question (ID: 9849)

Question:
$\text{A block is carried slowly up an inclined plane. If } W_f \text{ is work done by the friction, } W_N \text{ is work done by the reaction force, } W_g \text{ is work done by the gravitational force and } W_{ex} \text{ is the work done by an external force, then choose the correct relation(s):}$
Options:
  • 1. $W_N + W_f + W_g + W_{ex} = 0$
  • 2. $W_N = 0$
  • 3. $W_{ex} + W_f = -W_g$
  • 4. $\text{All of these}$
Solution:
\text{Analysis of each statement:} \text{1. The normal force is perpendicular to displacement, so } W_N = 0. \text{2. "Carried slowly" means } \Delta KE = 0. \text{ By the Work-Energy Theorem, } W_{\text{net}} = W_N + W_f + W_g + W_{\text{ex}} = \Delta KE = 0. \text{3. From } W_N + W_f + W_g + W_{\text{ex}} = 0 \text{ and } W_N = 0, \text{ it follows that } W_{\text{ex}} + W_f + W_g = 0, \text{ which can be rewritten as } W_{\text{ex}} + W_f = -W_g. \text{Since all three statements are correct, the answer is All of these.}

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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
}