Answer:

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 :

mg=kx………….(1)

For the final condition :

(m+m')g=k(x+10)……..(2)

Solving equation (1) and (2)

dfrac{mg}{(m+m')g}=dfrac{x}{x+0.1}

dfrac{x}{x+0.1}=dfrac{0.005}{(0.005+0.115)}

x = 0.0042 m

Now put the value of x in equation (1)

k=dfrac{mg}{x}

k=dfrac{0.005times 9.79}{0.0042}

k=11.65 N/m

Hence, this is the required solution.