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QJEEM21Ph7a55: JEE Main 2021 Physics - Class 11 JEEM21Ph7a55 Solution: A block of mass m slides along a floor while a force of…

A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ\theta as shown in figure. The coefficient of kinetic friction is μ\mu k. then, the block's acceleration 'a' is given by :

(g is acceleration due to gravity)

JEE Main 2021 (Online) 16th March Morning Shift Physics - Laws of Motion Question 84 English

Multiple Choice Options

  • A Fmcos⁡θ+μK(g−Fmsin⁡θ){F \over m}\cos \theta + {\mu _K}\left( {g - {F \over m}\sin \theta } \right)
  • B Fmcos⁡θ−μK(g−Fmsin⁡θ){F \over m}\cos \theta - {\mu _K}\left( {g - {F \over m}\sin \theta } \right)
  • C −- Fmcos⁡θ−μK(g−Fmsin⁡θ){F \over m}\cos \theta - {\mu _K}\left( {g - {F \over m}\sin \theta } \right)
  • D Fmcos⁡θ−μK(g+Fmsin⁡θ){F \over m}\cos \theta - {\mu _K}\left( {g + {F \over m}\sin \theta } \right)

Step-by-Step Verified Solution

Drawing the FBD of the block.

JEE Main 2021 (Online) 16th March Morning Shift Physics - Laws of Motion Question 84 English Explanation

⇒\Rightarrow N = mg −- Fsin θ\theta ..... (1)

Also, Fcos θ\theta −- μ\mu kN = m . a ..... (2)

Substituting the value of N from eq. (1) in eq. (2)

⇒\Rightarrow Fcos θ\theta −- μ\mu k(mg −- Fsin θ\theta ) = m . a

⇒\Rightarrow a = Fm{F \over m} cos θ\theta −- μ\mu k (g−Fmsin⁡θ)\left( {g - {F \over m}\sin \theta } \right)

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