Two blocks A and B of same masses attatched with a light spring are su

1.	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» 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 )

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»

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

→ Acceleration of block "A" and "B" .

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)

→ T - mg - mg = 0

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 ,

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

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

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

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