A spring with a constant of $5 \frac{k g}{s^{2}}$ $5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $6 k g$ $6 kg$ and speed of $8 \frac{m}{s}$ $8 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-12-13T07:14:15

The compression is $= 8.76 m$ $=8.76m$

Mass of the object is $m = 6 k g$ $m=6kg$

Speed of the object is $v = 8 m s^{- 1}$ $v=8ms^-1$

The kinetic energy of the object is

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 6 \cdot 8^{2} = 192 J$ $KE=1/2mv^2=1/2*6*8^2=192J$

This energy will be stored in the spring

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

The spring constant is $k = 5 k g s^{- 2}$ $k=5kgs^-2$

Let the compression of the spring be $= x m$ $=xm$

Therefore,

$192 = \frac{1}{2} \cdot 5 \cdot x^{2}$ $192=1/2*5*x^2$

$x^{2} = \frac{2 \cdot 192}{5} = 76.8 m^{2}$ $x^2=(2*192)/5=76.8m^2$

The compression is

$x = \sqrt{76.8} = 8.76 m$ $x=sqrt76.8=8.76m$

A spring with a constant of 5 (kg)/s^25kgs25 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 6 kg6kg6 kg and speed of 8 m/s8ms8 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?