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?

Question

1901 views around the world

You can reuse this answer
Creative Commons License

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#

enter image source here

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#/