Amass of 30.0 grams hangs at rest from the lower end of a long vertical spring. you add different amounts of additional mass δm to the end of the spring and measure the increase in length δl of the spring due to the additional mass. you plot your data in the form δl (vertical axis) versus δm (horizontal axis). plotted this way, your data lies very close to a straight line that has slope 0.0520 m/kg. what is the force constant k of the spring? use g = 9.80 m/s2.

k=188.46 N/m

Explanation:

Given that

m= 30 gm

Additional mass = ΔM

Lets take spring coefficient = k

at the equilibrium position

F= k  Δl =( m +  ΔM)g

Δl = change in the spring length

k  Δl = ( m +  ΔM)g

Δl = m g /k + g/k ΔM                     1

We know that equation of line given as

y = c + S x                         2

By compare equation 2 and 1

slope = s

S= g/k

Given that slope S= 0.052 m/kg

S =g/ k= 0.052

g=9.8 m/s²

9.8 = 0.052 k

k=188.46 N/m

So the spring constant is 188.46 N/m