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Current Question (ID: 11286)

Question:

$\text{A block } \text{P} \text{ of mass } m \text{ is placed on a frictionless horizontal surface. Another block } \text{Q} \text{ of same mass is kept on } \text{P} \text{ and connected to the wall with the help of a spring of spring constant } k \text{ as shown in the figure. } \mu_{s} \text{ is the coefficient of friction between } \text{P} \text{ and } \text{Q}. \text{ The blocks move together performing } \text{SHM} \text{ of amplitude } \text{A}. \text{ The maximum value of the friction force between } \text{P} \text{ and } \text{Q} \text{ will be:}$

Options:

1. $k\text{A}$
2. $\frac{k\text{A}}{2}$ (Correct)
3. $\text{zero}$
4. $\mu_{s}mg$

Solution:

$\text{When two blocks perform simple harmonic motion together, then at the extreme position (at amplitude } = \text{A}), \text{the restoring force on the system of two blocks is } \text{F} = k\text{A}. \text{According to Newton's second law, } \text{F} = (m+m)a = 2ma. \text{Thus, } k\text{A} = 2ma \Rightarrow a = \frac{k\text{A}}{2m}. \text{For the block } \text{P} \text{ to move along with block } \text{Q}, \text{ the force required on block } \text{P} \text{ is provided by the friction between the blocks. The maximum force needed to keep block } \text{P} \text{ accelerating with the system is the pseudo force on block } \text{P}, \text{ which is equal to } m\text{a}. \text{The pseudo force on block } \text{P} \text{ is } m\text{a} = m(\frac{k\text{A}}{2m}) = \frac{k\text{A}}{2}. \text{This force must be less than or equal to the maximum static friction, but for the maximum possible friction, the pseudo force is equal to it. Therefore, the maximum friction force required is } \frac{k\text{A}}{2}.$

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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
}