Two masses, M and m are connected together by a pulley two strings and a stretched spring of force constant k as shown. Assume that string, pulley and spring all are massless and surface below m is smooth. The amount by which the spring is stretched is :(assume system moving with constant acceleration)

Question

Two masses, M and m are connected together by a pulley two strings and a stretched spring of force constant k as shown. Assume that string, pulley and spring all are massless and surface below m is smooth. The amount by which the spring is stretched is :(assume system moving with constant acceleration) (A) ( mMg / k ( m + M )) (B) (2 m M g/k(m+M)) (C) ( Mg / k ) (D) (( m + M ) g /2 k )

Tardigrade Admin · Accepted Answer

Acceleration of blocks <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ea69ad3184ff6bb3cc45dfb2258f19f3-.png" /> a=(M g/m+M) Tension in the string/spring T = ma =( mMg / m + M ) ∴ Extension in the string, x =( T / k ) =( mMg / k ( m + M ))

$\frac{m M g}{k ( m + M )}$

$\frac{2 m M g}{k ( m + M )}$

$\frac{M g}{k}$

$\frac{( m + M ) g}{2 k}$

Acceleration of blocks

$a = \frac{M g}{m + M}$
Tension in the string/spring
$T = ma = \frac{m M g}{m + M}$
$∴$ Extension in the string, $x = \frac{T}{k}$
$= \frac{m M g}{k ( m + M )}$

k(m+M)mMg​

k(m+M)2mMg​

kMg​

2k(m+M)g​

Acceleration of blocks a=m+MMg​ Tension in the string/spring T=ma=m+MmMg​ ∴ Extension in the string, x=kT​ =k(m+M)mMg​

$\frac{m M g}{k ( m + M )}$

$\frac{2 m M g}{k ( m + M )}$

$\frac{M g}{k}$

$\frac{( m + M ) g}{2 k}$

Acceleration of blocks

$a = \frac{M g}{m + M}$
Tension in the string/spring
$T = ma = \frac{m M g}{m + M}$
$∴$ Extension in the string, $x = \frac{T}{k}$
$= \frac{m M g}{k ( m + M )}$