A spring with a constant of $9$ $9$ $k g s^{- 2}$ $kgs^-2$ is lying on the ground with one end attached to a wall. An object with a mass of $2$ $2$ $k g$ $kg$ and speed of $12$ $12$ $m s^{- 1}$ $ms^-1$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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A spring with a constant of $9$ $9$ $k g s^{- 2}$ $kgs^-2$ is lying on the ground with one end attached to a wall. An object with a mass of $2$ $2$ $k g$ $kg$ and speed of $12$ $12$ $m s^{- 1}$ $ms^-1$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-11T07:54:32

The compression is $= 5.66 m$ $=5.66m$

Explanation:

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The spring constant is $k = 9 k g s^{- 2}$ $k=9kgs^-2$

The kinetic energy of the object is

$K E = \frac{1}{2} m u^{2}$ $KE=1/2m u^2$

$K E = \frac{1}{2} \cdot 2 \cdot {(12)}^{2} = 144 J$ $KE=1/2*2*(12)^2=144J$

This kinetic energy will be stored in the spring as potential energy.

$P E = 144 J$ $PE=144J$

So,

$\frac{1}{2} k x^{2} = 144$ $1/2kx^2=144$

$x^{2} = 2 \cdot \frac{144}{9} = 32 m^{2}$ $x^2=2*(144)/(9)=32m^2$

$x = \sqrt{32} = 5.66 m$ $x=sqrt(32)=5.66m$

Answer 2 · 2017-06-11T07:56:44

The displacement of the spring, $x$ $x$ , will be $4$ $4$ $m$ $m$ .

Explanation:

The kinetic energy of the moving object will be $E_{k} = \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times 12^{2} = 144$ $E_k=1/2mv^2=1/2xx2xx12^2=144$ $J$ $J$ .

This will be converted to spring potential energy: $E_{sp} = \frac{1}{2} k x^{2}$ $E_"sp"=1/2kx^2$ .

Rearranging: $x = \sqrt{\frac{E_{sp}}{k}}$ $x=sqrt((E_"sp")/k)$

We can substitute in the calculated value of $E_{k}$ $E_k$ for $E_{sp}$ $E_"sp"$ :

$x = \sqrt{\frac{E_{sp}}{k}} = \sqrt{\frac{E_{k}}{k}} = \sqrt{\frac{144}{9}} = \frac{12}{3} = 4$ $x=sqrt((E_"sp")/k)=sqrt((E_k)/k)=sqrt((144)/9) = 12/3 = 4$ $m$ $m$

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A spring with a constant of $9$ $9$ $k g s^{- 2}$ $kgs^-2$ is lying on the ground with one end attached to a wall. An object with a mass of $2$ $2$ $k g$ $kg$ and speed of $12$ $12$ $m s^{- 1}$ $ms^-1$ collides with and compresses the spring until it stops moving. How much will the spring compress?

2 Answers

Explanation:

Explanation:

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A spring with a constant of 99 kgs−2kgs^-2 is lying on the ground with one end attached to a wall. An object with a mass of 22 kgkg and speed of 1212 ms−1ms^-1 collides with and compresses the spring until it stops moving. How much will the spring compress?

2 Answers

Explanation:

Explanation:

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A spring with a constant of $9$ $9$ $k g s^{- 2}$ $kgs^-2$ is lying on the ground with one end attached to a wall. An object with a mass of $2$ $2$ $k g$ $kg$ and speed of $12$ $12$ $m s^{- 1}$ $ms^-1$ collides with and compresses the spring until it stops moving. How much will the spring compress?