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Current Question (ID: 9762)

Question:

$\text{A body of mass } m \text{ is kept on a rough horizontal surface (coefficient of friction = } \mu \text{). A horizontal force is applied to the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by } \vec{F} \text{ where:}$

Options:

1. $|\vec{F}| = mg + \mu mg$
2. $|\vec{F}| = \mu mg$
3. $|\vec{F}| \leq mg\sqrt{1 + \mu^2}$
4. $|\vec{F}| = mg$

Solution:

$\text{Hint: Recall the concept of static friction.}$ $\text{Step: Find the resultant of friction and normal reaction.}$ $\text{The resultant of the normal reaction } N \text{ and the frictional force } f_s \text{ acting on the body.}$ $\text{Since the surface is horizontal and no vertical forces are applied, the normal force balances the weight:}$ $N = mg$ $\text{The static friction force opposes the applied force and adjusts itself to prevent motion.}$ $f_s \leq \mu N = \mu mg$ $\text{Since the body is not moving, the friction force } f_s \text{ is exactly equal to the applied force } F.$ $\text{Now balance the horizontal and vertical forces:}$ $N = mg$ $F_{ext} = f \leq \mu mg$ $\text{The normal force } N \text{ and the frictional force } f_s \text{ act perpendicular to each other.}$ $\text{The resultant } R \text{ of these two perpendicular forces can be found using the Pythagorean theorem:}$ $R = \sqrt{N^2 + f_s^2}$ $R = \sqrt{N^2 + F^2} \leq mg\sqrt{1 + \mu^2}$ $\text{Thus the resultant of the normal reaction and the frictional force is}$ $|\vec{F}| \leq mg\sqrt{1 + \mu^2}$ $\text{Hence, option (3) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}