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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2021 · 16 March · Morning Shift · Physics · Laws of Motion

A block of mass m slides along a floor while a force of magnitude F...

A block of mass mm slides along a floor while a force of magnitude FF is applied to it at an angle θ\theta as shown in figure. The coefficient of kinetic friction is μK{\mu}_{K}. Then, the block's acceleration aa is given by : (gg is acceleration due to gravity)

A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ as shown in figure.

Options

  1. A−Fmcosθ−μK(g−Fmsinθ)- \dfrac{F}{m}cos\theta - {\mu}_{K}\left(g - \dfrac{F}{m}sin\theta\right)
  2. BFmcosθ−μK(g−Fmsinθ)\dfrac{F}{m}cos\theta - {\mu}_{K}\left(g - \dfrac{F}{m}sin\theta\right)
  3. CFmcosθ−μK(g+Fmsinθ)\dfrac{F}{m}cos\theta - {\mu}_{K}\left(g + \dfrac{F}{m}sin\theta\right)
  4. DFmcosθ+μK(g−Fmsinθ)\dfrac{F}{m}cos\theta + {\mu}_{K}\left(g - \dfrac{F}{m}sin\theta\right)

Correct answer

B. Fmcosθ−μK(g−Fmsinθ)\dfrac{F}{m}cos\theta - {\mu}_{K}\left(g - \dfrac{F}{m}sin\theta\right)

Step-by-step solution

A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ as shown in figure.

N=mg−fsinθN = mg - fsin\theta

Fcos\theta - {\mu}_{k}N = ma</p><p>Fcosθ−μk(mg−Fsinθ)=ma</p><p>Fcos\theta - {\mu}_{k}\left(mg - Fsin\theta\right) = ma

a=Fmcosθ−μK(g−Fmsinθ)a = \dfrac{F}{m}cos\theta - {\mu}_{K}\left(g - \dfrac{F}{m}sin\theta\right)

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