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Current Question (ID: 9854)

Question:

$\text{A block of mass } m \text{ is connected to a spring of force constant } K\text{. Initially, the block is at rest and the spring is relaxed. A constant force } F \text{ is applied horizontally towards the right. The maximum speed of the block will be:}$

Options:

1. $\frac{F}{\sqrt{2mK}}$
2. $\frac{\sqrt{2}F}{\sqrt{mK}}$
3. $\frac{F}{\sqrt{mK}}$
4. $\frac{2F}{\sqrt{2mK}}$

Solution:

$\text{Hint: At maximum speed, the net force on the block is zero.}$ $\text{Step 1: Draw FBD of the block at the mean position.}$ $\text{Step 2: Find the mean position value when the block will have maximum speed.}$ $F = kx_0$ $\Rightarrow x_0 = \frac{F}{k}$ $\text{Step 3: Find the velocity of the block.}$ $W_{\text{spring}} + W_{\text{ext}} = \Delta K$ $\Rightarrow \int_0^{x_0} -kx dx + Fx_0 = \frac{1}{2}mv^2$ $\Rightarrow \frac{1}{2}mv^2 = -\frac{1}{2}kx_0^2 + Fx_0$ $\Rightarrow \frac{1}{2}mv^2 = \frac{F^2}{2K} \left[x_0 = \frac{F}{k}\right]$ $\Rightarrow v^2 = \frac{F^2}{mK}$ $\Rightarrow v = \frac{F}{\sqrt{mK}}$ $\text{Hence, option (3) is the correct answer.}$

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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
}