A50 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. what is the spring constant of the spring (in n/m)? (assume g = 9.79 m/s2.

The spring constant of the spring is 11.65 N/m

Explanation:

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

Where

k is the spring constant

For the initial condition :

.............(1)

For the final condition :

........(2)

Solving equation (1) and (2)

x = 0.0042 m

Now put the value of x in equation (1)

Hence, this is the required solution.

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.