Question

A mass 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 0.0268 m/kg. What is the force constant k of the spring? Use g = 9.80 m/s2.

Answer 1

Answer:

366N/m

Explanation:

According to Hooke's law:

$F=-kx$

avoiding the minus symbol and substituting the force to the gravitational force, we have:

$m.g=k.x\\k=\frac{m.g}{x}=\frac{m}{x}*g\\k=\frac{1}{s}*g\\k=\frac{1}{0.0268m/kg}*9.80m/s^2\\\\k=366N/m$

The spring constant is 366N/m

Answer 2

Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0     =>    Fe = m*g

Using Hook's law:

K*X = m*g    Solving for K:

K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

K = 10 / 0.0268 = 373.13N/m