A spring with a constant of #6/7 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #3/5 kg# and speed of #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(189/480) m#

Kinetic energy of object #= (mv^2)/2 = (3/5 kg × (3/4 ms^-1)^2) / 2 = 27/160 J#

Potential energy stored in spring #= (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

#27/160 J = (3x^2)/7 kg . s^-2#

#x = sqrt((7 × 27/160) / 3) m = sqrt(189/480) m#

Answer 2 · 2017-10-18T15:14:27

The compression is #=0.63m#

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The spring constant is #k=6/7kgs^-2#

The kinetic energy of the object is

#KE=1/2m u^2#

The mass is #m=3/5kg#

The speed is #u=3/4ms^-1#

#KE=1/2*3/5*(3/4)^2=27/160J#

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

#PE=27/160J#

The spring constant is #=6/7kgs^-2#

So,

#1/2kx^2=27/160#

#x^2=2*(27/160)/(6/7)=189/480m^2#

#x=sqrt(189/480)=0.63m#/