Question

A 0.20 kg mass (m1) hangs vertically from a spring and an elongation of the spring of 9.50 cm (r1) is recorded. With a mass (m2) of 1.00 kg hanging on the spring, a second elongation (r2) of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter (N/m). (Note: The equilibrium position is not zero.)

Answer 1

Answer:

k=320N/m

Explanation:

Step one:

given data

Let the initial/equilibrum position be x

mass m1= 0.2kg

F1= 0.2*10= 2N

elongation e= 9.5cm= 0.095m

mass m2=1kg

F2=1*10= 10N

elongation e= 12cm= 0.12m

Step two:

From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F

F=ke

2=k(0.095-a)

2=0.095k-ka----------1

10=k(0.12-a)

10=0.12k-ka----------2

solving equation 1 and 2 simultaneously

 

   10=0.12k-ka----------2

-   2=0.095k-ka----------1

   8=0.025k-0

divide both side by 0.025

k=8/0.025

k=320N/m