Search any question & find its solution
Question:
Answered & Verified by Expert
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, $\left.g=10 \mathrm{~ms}^{-2}\right)\left(\sin 37^{\circ}=\frac{3}{5}\right)$
Options:
Solution:
1548 Upvotes
Verified Answer
The correct answer is:
$8 \mathrm{~ms}^{-2}$
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, $F_{\text {net }}=F_V-\left(F_r+w\right)=m a$
$$
\begin{array}{ll}
\therefore & F_{\text {net }}=54-18-20=16 \mathrm{~N}=m a \\
\Rightarrow & a=\frac{16}{2}=8 \mathrm{~m} / \mathrm{s}^2
\end{array}
$$
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, $F_{\text {net }}=F_V-\left(F_r+w\right)=m a$
$$
\begin{array}{ll}
\therefore & F_{\text {net }}=54-18-20=16 \mathrm{~N}=m a \\
\Rightarrow & a=\frac{16}{2}=8 \mathrm{~m} / \mathrm{s}^2
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.