Question

In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the spring of 9.50 cm is recorded. With 1.00 kg hanging on the spring, a second elongation 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

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

$k (9.5-l)=0.2*g$

$k (9.5-l)=0.2*9.8$

For state 2 we have that with 1Kg there is an elongation of 12cm

$k (12-l)= 1*g$

$k (12-l)= 1*9.8$

We have two equations with two unknowns therefore solving for both,

$k = 3.136N/cm$

$l = 8.877cm$

In this way converting the units,

$k = 3.136N/cm(\frac{100cm}{1m})$

$k = 313.6N/m$

Therefore the spring constant is 313.6N/m