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Current Question (ID: 19263)

Question:

$\text{A constant force acts on a body of mass } 1 \text{ kg, providing it a kinetic energy of } 1800 \text{ J by the end of } 5^{\text{th}} \text{ second. If the body was initially at rest, at the beginning of the action of force, then the magnitude of the force is equal to:}$

Options:

1. $30 \text{ N}$
2. $12 \text{ N}$
3. $25 \text{ N}$
4. $15 \text{ N}$

Solution:

$\text{Hint: } a = \frac{F}{m}$ $\text{Step 1: Find the displacement of the block.}$ $\text{By using second equation of motion we get;}$ $s = ut + \frac{1}{2} at^2 = \frac{1}{2} at^2 \quad [u = 0]$ $\Rightarrow s = \frac{1}{2} \frac{F}{m} t^2 \quad [F = ma]$ $\text{Step 2: Find the magnitude of the force.}$ $\text{According to the work-energy theorem;}$ $W = \Delta K. E$ $\Rightarrow W = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 = \frac{1}{2} mv^2 \quad [u = 0]$ $\Rightarrow Fs = 1800 \text{ J} \quad [W = Fs]$ $\Rightarrow \frac{F^2 t^2}{2m} = 1800 \text{ J}$ $\Rightarrow F = \sqrt{\frac{2 \times 1 \times 1800}{25}} = \sqrt{144} = 12 \text{ N}$ $\text{Therefore, the magnitude of the force is } 12 \text{ N.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}