Question

Two blocks A and B of same masses attatched with a light spring are suspended by a string as shown in figure .Find the acceleration of block A and block B just after cut the string .​

Answer 1

Answer :-

$\implies \boxed{ \sf{a_{A} = 2g \: or \: 20 \: \frac{m} {{s}^{2} } }} \\ \implies \boxed{ \sf{ a_{B} = 0}} \:$

To Find :-

→ Acceleration of block "A" and "B" .

Explanation:-

According to the question -

→ Mass of both blocks is same .

(1) Before cut the string ( refer to attachment )

Before cut the string tension (T) , spring force (kx) and gravitational force (mg) is working on block "A" ,

Before cut the string there is no acceleration in both blocks .

hence ,

→ T - kx - mg = 0 .....eq.(1) ( on block A)

and ,

→ kx - mg = 0 ( on block B)

→ kx = mg [ put in eq.(1) ]

→ T - mg - mg = 0

→ T = 2mg

now,

(2) Just after cut the string ( refer to the attachment )

After cutting the string these is Acceleration in both block .( After cut the string Tension is removed on block "A" )

hence ,

$\sf{ Let \:acceleration\:of\:block\:A\:is\:a_{A}}\\ \sf{and\:of\:B\:is\:a_{B}}$

• On block "B"

$\implies \sf{ kx - mg = ma_{B} } \\ \\ \implies \sf{mg - mg =m a_{B} } \\ \\ \implies \boxed{ \sf{ a_{B} = 0}}$

• On block "A"

$\implies \sf{ mg + kx = ma_{A}} \\ \\ \implies \sf{ma_{A} = 2mg} \\ \\ \implies \boxed{ \sf{a_{A} = 2g \: or \: 20 \: \frac{m} {{s}^{2} } }}$

Note - After cut the string these is no normal contact force .( so neglect it )

Answer 2

$\tt{\huge{\red{Hello!!! }}}$

Question :

Two blocks A and B of same masses attatched with a light spring are suspended by a string as shown in figure .Find the acceleration of block A and block B just after cut the string .

Solution :

$\tt{\orange {Step-By-Step-Explaination \::- }}$

See the following attachment...

The image shows the F.B.D diagram.

Referring to the diagram,

We can derive the following equations :

$\tt{ \purple{ \leadsto{mg \: - \: kx \: = 0 \: }}}.......(1)$

$\tt{ \orange{ \leadsto{kx \: + \: mg \: - \: t \: = 0 \: }}}......(2)$

Note :

After the string is cut, the tension will become zero but the spring force i.e, kx will remain same.

This is a crucial point to note while solving.

Now draw the F.B.D again and solve.

Equations :

$\tt{ \purple{ \leadsto{mg \: = \: kx \: \: }}}.......(3)$

$\tt{ \purple{ \leadsto{mg \: - \: kx \: = ma_{b} \: }}}.......(4) \\ \\ = > a_{b }\: = 0$

$\tt{ \blue{ \leadsto{mg \: + \: kx \: = ma_{a} \: }}}.......(4) \\ \\ = > a_{a}\: = \: 2g$

Therefore the :

• acceleration of the block A = 2g

• acceleration of the block B = 0.