A spring with a constant of $\frac{6}{7} \frac{k g}{s^{2}}$ $6/7 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $\frac{3}{5} k g$ $3/5 kg$ and speed of $\frac{3}{4} \frac{m}{s}$ $3/4 m/s$ 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 $\frac{6}{7} \frac{k g}{s^{2}}$ $6/7 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $\frac{3}{5} k g$ $3/5 kg$ and speed of $\frac{3}{4} \frac{m}{s}$ $3/4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-10-18T14:13:34

$\sqrt{\frac{189}{480}} m$ $sqrt(189/480) m$

Explanation:

Kinetic energy of object $= \frac{m v^{2}}{2} = \frac{\frac{3}{5} k g \times {(\frac{3}{4} m s^{- 1})}^{2}}{2} = \frac{27}{160} J$ $= (mv^2)/2 = (3/5 kg × (3/4 ms^-1)^2) / 2 = 27/160 J$

Potential energy stored in spring $= \frac{k x^{2}}{2} = \frac{\frac{6}{7} k g s^{- 2} \times x^{2}}{2} = \frac{3 x^{2}}{7} k g . s^{- 2}$ $= (kx^2)/2 = (6/7 kgs^-2 × x^2) / (2) = (3x^2)/7 kg.s^-2$

According to law of Conservation of Energy
Total energy of object = Total energy absorbed by spring

$\frac{27}{160} J = \frac{3 x^{2}}{7} k g . s^{- 2}$ $27/160 J = (3x^2)/7 kg . s^-2$

$x = \sqrt{\frac{7 \times \frac{27}{160}}{3}} m = \sqrt{\frac{189}{480}} m$ $x = sqrt((7 × 27/160) / 3) m = sqrt(189/480) m$

Answer 2 · 2017-10-18T15:14:27

The compression is $= 0.63 m$ $=0.63m$

Explanation:

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The spring constant is $k = \frac{6}{7} k g s^{- 2}$ $k=6/7kgs^-2$

The kinetic energy of the object is

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

The mass is $m = \frac{3}{5} k g$ $m=3/5kg$

The speed is $u = \frac{3}{4} m s^{- 1}$ $u=3/4ms^-1$

$K E = \frac{1}{2} \cdot \frac{3}{5} \cdot {(\frac{3}{4})}^{2} = \frac{27}{160} J$ $KE=1/2*3/5*(3/4)^2=27/160J$

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

$P E = \frac{27}{160} J$ $PE=27/160J$

The spring constant is $= \frac{6}{7} k g s^{- 2}$ $=6/7kgs^-2$

So,

$\frac{1}{2} k x^{2} = \frac{27}{160}$ $1/2kx^2=27/160$

$x^{2} = 2 \cdot \frac{\frac{27}{160}}{\frac{6}{7}} = \frac{189}{480} m^{2}$ $x^2=2*(27/160)/(6/7)=189/480m^2$

$x = \sqrt{\frac{189}{480}} = 0.63 m$ $x=sqrt(189/480)=0.63m$ /

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