A spring with a constant of #3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #4 kg# and speed of #6 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 #3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #4  kg# and speed of #6  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Layla Ahmed · Accepted Answer

By conservation of mechanical energy when  the spring  is compressed fully after collision 
PE gained by the spring = Initial KE of the colliding object
#=>cancel (1/2)kx^2=cancel(1/2)mv^2#
where 
m = mass=4kg
v=velocity of the object=#6m/s#
k = force constant=#3(kg)/s^2#
x= compression of spring=?
#=>x^2=mv^2/k#
#=>x=sqrt(m/kv^2)=sqrt(m/k)xxv=sqrt((4kg)/(3kg)/s^2)xx6m/s=6.9m#

Olivia Askew · Answer

undefined To find the compression of the spring, you can use the conservation of mechanical energy formula: $ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 $, where $ k $ is the spring constant, $ x $ is the compression of the spring, $ m $ is the mass of the object, and $ v $ is the velocity of the object.

Plugging in the given values: $ k = 3 \, \text{kg/s}^2 $, $ m = 4 \, \text{kg} $, and $ v = 6 \, \text{m/s} $,

$ \frac{1}{2}(3 \, \text{kg/s}^2)x^2 = \frac{1}{2}(4 \, \text{kg})(6 \, \text{m/s})^2 $

$ 3x^2 = 4 \times 36 $

$ 3x^2 = 144 $

$ x^2 = \frac{144}{3} $

$ x^2 = 48 $

$ x = \sqrt{48} $

$ x \approx 6.93 \, \text{m} $

Therefore, the spring will compress approximately $ 6.93 \, \text{m} $.