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Current Question (ID: 19275)

Question:

$\text{A } 500 \text{ kg object slides across a horizontal surface where the coefficient of kinetic friction is } \mu = 0.7.$ $\text{To maintain a constant velocity of } 10 \text{ m/s, an external force is applied horizontally in the direction of motion.}$ $\text{The object travels a total distance of } 4 \text{ km under these conditions.}$ $\text{The work done by the external force is:}$

Options:

1. $3.5 \times 10^6 \text{ J}$
2. $28 \times 10^6 \text{ J}$
3. $7 \times 10^6 \text{ J}$
4. $14 \times 10^6 \text{ J}$

Solution:

$\text{Hint: } W = \vec{F} \cdot \vec{s}$ $\text{Step: Find the work done in moving a body.}$ $\text{As the body is moving with constant velocity then, the change in kinetic energy is zero i.e., } \Delta K.E = 0.$ $\text{The force due to friction is given by:}$ $f = \mu mg$ $\Rightarrow f = 0.7 \times 500 \times 10$ $\Rightarrow f = 3500 \text{ N}$ $\text{According to the work-energy theorem;}$ $W + W_f = \Delta K.E$ $\Rightarrow W = -W_f$ $\text{The work done due to friction is given by;}$ $W = -W_f = -(-fs) \text{ [Work done by friction is negative]}$ $W = fs$ $W = 3500 \times 4000$ $W = 14 \times 10^6 \text{ J}$ $\text{Hence, option (4) is the correct answer.}$

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
}