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 $8 k g$ $8 kg$ and speed of $2 \frac{m}{s}$ $2 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-05-17T06:02:36

The compression is $= 2.14 m$ $=2.14m$

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The spring constant is $k = 7 k g s^{- 2}$ $k=7kgs^-2$

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 8 \cdot {(2)}^{2} = 16 J$ $KE=1/2*8*(2)^2=16J$

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

$P E = 16 J$ $PE=16J$

So,

$\frac{1}{2} k x^{2} = 16$ $1/2kx^2=16$

$x^{2} = 16 \cdot \frac{2}{7} = 4.57 m^{2}$ $x^2=16*2/(7)=4.57m^2$

$x = \sqrt{4.57} = 2.14 m$ $x=sqrt(4.57)=2.14m$

A spring with a constant of 7 (kg)/s^27kgs27 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8 kg8kg8 kg and speed of 2 m/s2ms2 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?