A spring with a constant of #6 (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 #1 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

1 Answer
Mar 15, 2016

Answer:

Compression of the spring #=0.82m# rounded to second place of decimal.

Explanation:

Suppose the spring be compressed by #X# meter.

Force which compressed the string is the force applied by the moving body which came to rest after acting on the spring.
The speed of the object changed from #1ms^-1# to zero.
Using the expression from linear motion
#v^2-u^2=2as#,
where #v# is final velocity, #u# initial velocity, #a# acceleration and #s# distance moved.
Inserting the given values we obtain

#0^2-1^2=2aX#,

recall that the compressing objects moves distance #X# before it came to stop. This is the amount of compression of the spring produced by it. Solving for acceleration we obtain
#a=-1/(2X)#, #-ve# sign indicates that the acceleration is in a direction opposite to direction of movement of object or it is retardation.

We know that Force #F=ma#
#:. F=8xx(-1/(2X))=-4/X#

From Hooke's law for springs we know that
Reaction Force #F=-kX#, where #k# is the spring constant.
Now inserting the various values in the equation we obtain
#-4/X=-6X#, solving for #X#
#X^2=4/6#
or #X=sqrt(4/6)=0.82m# rounded to second place of decimal.