Question

In the given figure, spring, string and pulley are massless. System released from rest
when spring in its natural length. Find maximum elongation in the spring.
​

Answer 1

Given:

spring, string and pulley are massless.

m is the mass of block and K is the spring constant of spring.

System is released from rest when spring is in its natural length.

To Find:

Maximum elongation of spring

Solution:

Free body diagram of mass m

1) mg - downwards

2)T - upwards

Net force = 0( system is in equilibrium.)

T = mg

Free body diagram of pulley

1)  T - downwards

2) 2(T') - upwards

        net force = 0

        2(T')= T

                  $T' = \frac{T}{2}$

Free body diagram of spring

1) T' - downwards

2) spring force Kx upwards

where x is the maximum elongation of spring

$T' = Kx\\x = \frac{T'}{K}$

substitute T' = 0.5 T = .5(mg)

$x = \frac{0.5(mg)}{K} \\x= \frac{mg}{2K}$

Maximum elongation of spring is   $x = \frac{mg}{2K}$