Q126 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Physics

A block of mass $2 \mathrm{~kg}$ is being pushed against a wall by a force $F=90 \mathrm{~N}$ as shown in the figure. If the coefficient of friction is 0.25 , then the magnitude of acceleration of the block is (Take, $\left.g=10 \mathrm{~ms}^{-2}\right)\left(\sin 37^{\circ}=\frac{3}{5}\right)$ Figure

  • A
    $16 \mathrm{~ms}^{-2}$
  • Correct answer
    B
    $8 \mathrm{~ms}^{-2}$
  • C
    $38 \mathrm{~ms}^{-2}$
  • D
    $54 \mathrm{~ms}^{-2}$

Solution

Block's weight (downward), $w=2 \times 10=20 \mathrm{~N}$ Vertical component of applied force (upwards), $$ F_V=F \sin \theta=90 \times 3 / 5=54 \mathrm{~N} $$ Maximum frictional force, $$ \begin{aligned} F_r & =\mu F \cos \theta=0 \cdot 25 \times 90 \times 4 / 5 \\ & =18 \mathrm{~N} \end{aligned} $$ $\therefore$ Net vertical force, $F_{\text {net }}=F_V-\left(F_r+w\right)=m a$ $$ \begin{array}{ll} \therefore & F_{\text {net }}=54-18-20=16 \mathrm{~N}=m a \\ \Rightarrow & a=\frac{16}{2}=8 \mathrm{~m} / \mathrm{s}^2 \end{array} $$

Hence, the correct answer is option: 'B'.

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