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 0.0364 m/kg. What is the force constant k of the spring? Use g = 9.80 m/s2.

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creamydhaka creamydhaka · Answer 1 · 2019-11-13T12:37:03+01:00

Answer:

[tex]k=269.231\,N.m^{-1}[/tex]

Explanation:

For a spring the elastic constant is the amount of force required for the deformation of unit length in the spring. It is usually denoted by k.

We have a relation between the force, length of deformation and elastic constant given as:

[tex]F=-k.\Delta x[/tex]

where:

[tex]\Delta x[/tex]= change in length

where the negative sign just denotes that the force is acting in opposite direction to the displacement.

Here we are given with the slope of Δx versus changing mass which is [tex]0.0364\,m.kg^{-1}[/tex].

For k we need the inverse of this value multiplied by the gravity to get the corresponding load of the given mass.

So,

[tex]k=9.8\times \frac{1}{0.0364}[/tex]

[tex]k=269.231\,N.m^{-1}[/tex]

abidemiokin abidemiokin · Answer 2 · 2021-11-11T16:34:53+01:00

The force constant k of the spring is 8.077N/m

According to hooke's law, the applied force on a spring is directly proportional to its extension. Mathematically,

F = ke

k is the force constant

e is the extension

Given the following parameters

extension e = slope = 0.0364m/kg

Force = mg = 0.03 × 9.8

F = 0.294N

Get the force constant "k"

k = F/e

k = 0.294/0.0364

k = 8.077N/m

Hence the force constant k of the spring is 8.077N/m

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