A spring with a constant of $7 \frac{k g}{s^{2}}$ $7 (kg)/(s^2)$ is lying on the ground with one end attached to a wall. An object with a mass of $4 k g$ $4 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ 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 $7 \frac{k g}{s^{2}}$ $7 (kg)/(s^2)$ is lying on the ground with one end attached to a wall. An object with a mass of $4 k g$ $4 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-01-20T04:37:31

6,8 m

Explanation:

The Kinetic energy of the object $E k = \frac{1}{2} m V^{2}$ $Ek = 1/2 mV^2$ becomes the elastic energy of the spring $E s = \frac{1}{2} m X^{2}$ $Es= 1/2 mX^2$ so
$m V^{2} = K X^{2}$ $mV^2 = K X^2$
$X = \sqrt{\frac{m}{k}} V$ $X= sqrt (m/k) V$
$X = \sqrt{\frac{4 k g}{7 k \frac{g}{s^{2}}}} 9 \frac{m}{s}$ $X= sqrt ((4kg)/(7kg/s^2)) 9 m/s$ = 6,8 m

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A spring with a constant of $7 \frac{k g}{s^{2}}$ $7 (kg)/(s^2)$ is lying on the ground with one end attached to a wall. An object with a mass of $4 k g$ $4 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

1 Answer

Explanation:

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A spring with a constant of 7kgs27 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 4kg4 kg and speed of 9ms 9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

1 Answer

Explanation:

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Impact of this question

A spring with a constant of $7 \frac{k g}{s^{2}}$ $7 (kg)/(s^2)$ is lying on the ground with one end attached to a wall. An object with a mass of $4 k g$ $4 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?