A spring with a constant of #6/5 (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 #5/4 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

A spring with a constant of #6/5 (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 #5/4  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Isabella Alvarez · Accepted Answer

#(5 sqrt(2))/8 "m"# Kinetic energy of object = Energy stored in spring #1/2mv^2 = 1/2kx^2# #x = vsqrt(m/k) = 5/4 "m/s"sqrt((3/5"kg")/(6/5 "kg/s"^2)) = 5/(4sqrt(2)) "m" = (5sqrt(2))/8 "m"#

Leo Abeyta · Answer

The compression is #=0.88m#

Mass of the object is #m=3/5kg#

Speed of the object is #v=5/4ms^-1#

The kinetic energy of the object is

#KE=1/2mv^2=1/2*3/5*(5/4)^2=0.46875J#

This energy will be stored in the spring

#E=1/2kx^2#

The spring constant is #k=6/5kgs^-2#

Let the compression of the spring be #=xm#

Therefore,

#0.46875=1/2*6/5*x^2#

#x^2=(5*0.46875)/3=0.78m^2#

The compression of the spring is

#x=sqrt0.78=0.88m#

Adam Arnold · Answer

undefined Use the conservation of energy principle. The spring's potential energy equals the initial kinetic energy of the object. The compression can be calculated using the formula: $ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 $. Solve for $ x $. $ x = \sqrt{\frac{m v^2}{k}} $. Substitute $ m = \frac{3}{5} \, \text{kg} $, $ v = \frac{5}{4} \, \text{m/s} $, and $ k = \frac{6}{5} \, \text{(kg)/s^2} $. Calculate to find $ x $.