A block is attached with an ideal spring and string, and is in equilibrium as shown in figure. The acceleration of the block just after breaking the string is $a_{1}$ and just after breaking the spring is $a_{2}$ . Then.

a_{1} = \frac{5 g}{8}

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a_{2} = \frac{3 g}{5}

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a_{1} = \frac{g}{2}

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a_{2} = g

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Solution

The correct option is A $a_{1} = \frac{5 g}{8}$
Balancing the forces
$T c o s 37 + k x c o s 37 = M g$
$k x s i n 37 = T s i n 37$
$T = k x 2 k x c o s 37 = M g$
$K x \frac{8}{5} = M g$
$x = \frac{5 M g}{8 k}$
Now if string is cut
$M g - K x * \frac{4}{5} = M a$
$M g - \frac{5 M g}{8 k} k \frac{4}{5} = M a$
$\frac{M g}{2} = M a$
$a = \frac{g}{2}$
$k x s i n 37 = M a_{x}$
$a_{x} = \frac{5 M g}{8} * \frac{3}{5}$
$a_{x} = \frac{3 M g}{8 M}$
$a_{1} = \sqrt{a_{y}^{2} + a_{x}^{2}}$
$= \frac{5 g}{8}$
If string is cut $T = k x$
Hence the acceleration will remain same
$a_{2} = \frac{5 g}{8}$

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Similar questions

Three blocks 1,2 and 3 are arranged with pulley and spring as shown in the figure. If string connecting blocks $m_{2}$ and $m_{3}$ is cut at point A, find the accelerations of masses $m_{1}$ , $m_{2}$ , and $m_{3}$ just after the string is cut at point A.

Q. Initially the system shown in figure is in equilibrium. At the moment the string is cut, the downward acceleration of blocks A and B are respectively

a_{1}

and

a_{2}

. The magnitudes of

a_{1}

and

a_{2}

are (Given acceleration due to gravity

= g

A block of mass 2 $M$ is attached to a massless spring with spring constant $k$ . This block is connected to two other blocks of masses $M$ and $2 M$ using two massless pulleys and strings. The accelerations of the blocks are $a_{1}, a_{2}$ and $a_{3}$ as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x_{0}$ . Which of the following option(s) is/are correct?
[ $g$ is the acceleration due to gravity. Neglect friction]