Question

The block shown in the figure is in equilibrium. Find acceleration of the block just after the string burns: The string is horizontal before burning.

• A. $\frac{3g}{5}$
• B. $\frac{4g}{5}$
• C. $\frac{4g}{3}$
• D. $Noneofthese$

Answer 1

Answer: (C)

$\frac{4g}{3}$

Free body diagram of the block will be as shown in fig.

Since block is in equillibrium, horizontal and vertical forces are balancing each other,

Vertically,

$F\cos {53}^{\circ }=mg$

$F\times \frac{3}{5}=mg$

$F=\frac{5}{3}mg$

When the string burned, Tension will be zero.

So let horizontal acceleration be $a$.

$\therefore ma=F\sin {53}^{\circ }$

$ma=\frac{5}{3}mg\times \frac{4}{5}$

$a=\frac{4}{3}g$