A spring extends in length by $4.0$ $4.0$ $c m$ $cm$ when a mass of $1.5$ $1.5$ $k g$ $kg$ is hung from it. What is the spring constant of the spring?

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A spring extends in length by $4.0$ $4.0$ $c m$ $cm$ when a mass of $1.5$ $1.5$ $k g$ $kg$ is hung from it. What is the spring constant of the spring?

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Answer 1 · 2017-06-07T12:11:40

$k = \frac{F_{x}}{x - x_{0}} = \frac{14.7 N}{0.04 m} = 367.5$ $k=(F_x)/(x-x_0)=(14.7" N")/(0.04" m")=367.5$ $N m^{- 1}$ $Nm^-1$

Explanation:

Let these variables represent the following

$F_{x} =$ $F_x=$ the force applied to the spring

$k =$ $k=$ the spring force constant

$x =$ $x=$ the distance from equilibrium

$x_{0} =$ $x_0=$ the spring equilibrium position

The equation to determine the spring constant, $k$ $k$ is

$k = \frac{F_{x}}{x - x_{0}}$ $k=(F_x)/(x-x_0)$

We are given the total measure of elongation is $4$ $4$ $c m$ $cm$ , which is the difference between $x$ $x$ and $x_{0}$ $x_0$ . That is,

$x - x_{0} = 4$ $x-x_0=4$ $c m = 0.04$ $cm=0.04$ $m$ $m$

We are also given the mass of the body, $1.5$ $1.5$ $k g$ $kg$ , and assuming this measurement is happening on Earth, we have the acceleration due to gravity $9.8$ $9.8$ ms $^{- 2}$ $""^(-2)$ .

$F_{x} = 9.8$ $F_x=9.8$ $(m s^{- 2}) \times 1.5$ $(ms^-2)xx1.5$ $k g = 14.7$ $kg=14.7$ $N$ $N$

Therefore, the final solution is

$k = \frac{F_{x}}{x - x_{0}} = \frac{14.7 N}{0.04 m} = 367.5$ $k=(F_x)/(x-x_0)=(14.7" N")/(0.04" m")=367.5$ $N m^{- 1}$ $Nm^-1$

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A spring extends in length by $4.0$ $4.0$ $c m$ $cm$ when a mass of $1.5$ $1.5$ $k g$ $kg$ is hung from it. What is the spring constant of the spring?

1 Answer

Explanation:

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A spring extends in length by $4.0$ $4.0$ $c m$ $cm$ when a mass of $1.5$ $1.5$ $k g$ $kg$ is hung from it. What is the spring constant of the spring?