Question

When a 400g mass is hung at the end of a vertical spring, the spring stretches 35cm. What is the spring constant of the spring, and how much further will it stretch if an additional 400g mass is hung from it?​

Answer 1

from Hoke's law,

Weight of body = spring restoring force

or, mg = Kx

or, K= mg/x = (400g × 980 cm/s²)/(35cm)

= 11200 Dyne/cm

= 11200 × 10^-5 N/(10^-2m)

= 11200 × 10^-3 N/m

= 11.2 N/m [ans]

again, weight body = mg + mg = 2mg

= 2 × 400 × 980 Dyne

restoring force, F' = Kx + Kx'

= K(35cm + x')

= 11200 Dyne/cm {35cm + x'}

so, 2 × 400 × 980 Dyne = 11200 Dyne/cm × {35cm + x'}

or, 70cm = 35cm + x'

or, x' = 35cm [Ans]