The spring constant of the spring is 11.65 N/m
It is given that, a 50 g mass hanger hangs motionless from a partially stretched spring. When a 65 gram mass is added to the hanger, the spring stretch increases by 10 cm.
Initial mass, m = 50 g = 0.005 kg
Final mass, m’ = 50 g + 65 g = 115 g = 0.115 kg
Initially, the elongation is x (say). The spring stretch increases by 10 cm or 0.1 m.
We know that,
F = k x
k is the spring constant
For the initial condition :
For the final condition :
Solving equation (1) and (2)
x = 0.0042 m
Now put the value of x in equation (1)
Hence, this is the required solution.