A spring with a constant of #1 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #8 kg# and speed of #9 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-05-03T09:44:10

#25.5m#

Conservation of energy tells us that there must be exactly the same amount of energy before a reaction as after.

Before the reaction energy is kinetic, since the object is moving, and energy is given by

#E=1/2mv^2#,

where #m# is mass and #v# is velocity.

After the reaction, the energy is potential, stored in the spring, and given by

#E=1/2kx^2#,

where #x# is the distance the spring compresses and #k# is the spring constant.

Since these two equations must be equal (conservation of energy above),

#1/2mv^2=1/2kx^2#

#therefore mv^2=kx^2#

The question is asking you about how much the spring compresses, so you rearrange to make #x# the subject.

#sqrt(mv^2/k)=x#.

Put in the values you know and solve

#sqrt((8*9^2)/1)=x#

#sqrt(648)=x=25.5m#