Question

In the figure shown, acceleration of the block is
(Take $g=10{\text{m/s}}^2$)

• A. $5.3{\text{m/s}}^2$
• B. zero
• C. $\left(\frac{50\sqrt{3}-4}{10}\right){\text{m/s}}^2$
• D. $(5\sqrt{3}-3){\text{m/s}}^2$

Answer 1

The correct option is A $5.3{\text{m/s}}^2$

FBD of $10\text{kg}$ block:


From FBD, we get
$R=10g\cos {60}^{\circ }-50\sin {53}^{\circ }\Rightarrow R=50-40=10\text{N}$

Maximum static friction $(f_s{)}_{max}={\mu }_{s}R=0.5\times 10=5\text{N}$

Kinetic friction $\left(f_k\right)={\mu }_{k}R=0.4\times 10=4\text{N}$
As, $10g\sin {60}^{\circ }>50\cos {53}^{\circ }$ the block will move down and force of friction acts upwards.
Then,
$10g\sin {60}^{\circ }-50\cos {53}^{\circ }-f_k=ma$
From the data given in the question,

$\Rightarrow a=\frac{10\times 10\times \frac{\sqrt{3}}{2}-[50\times \frac{3}{5}+4]}{10}$

$\Rightarrow a=\frac{50\sqrt{3}-34}{10}\approx 5.3{\text{m/s}}^2$