A spring is attached to the ceiling, and when a mass is suspended from the spring at rest, it stretches by 50 cm. If instead two identical copies of the same spring are attached to the mass and the ceiling, the springs will stretch by:

25 cm

Explanation:

Given that :

Displacement, x = 50 cm

In a spring, the force is given thus ;

F = kx ; where, F = Force, k = spring constant

Hence,

F = 50k - - - (1)

If two identical springs are attached to the mass ;

Fnet = F1 + F2

Fnet = F = kx' + kx'

x' = Displacement

Spring constant k and x' will be the same in the two springs ;

Hence,

F = 2kx' - - - (2)

Equating (1) and (2)

F = F

50k = 2kx'

50 = 2x'

x' = 50 / 2

x' = 25

Hence, the spring will stretch by 25cm

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