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 #9 kg# and speed of #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 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #9  kg# and speed of #4 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Isabella Alvarez · Accepted Answer

Compression: #x = \sqrt{m/k}v_0 = \sqrt{(9\quad kg)/(6\quad N/m)}(4\quad m/s) = 4.899\quad m#

Mechanical Energy Conservation: A system's mechanical energy is preserved when the forces operating on it are entirely conservative.

#\Delta E = \Delta K + \Delta U = 0#

#\DeltaE# - Change in total mechanical energy; #\DeltaK# - Change in kinetic energy; #\DeltaU# - Change in potential energy;

In this instance, the total mechanical energy of the system is maintained constant by increasing the potential energy in the spring by the same amount as the kinetic energy of the block decreases.

#m# - mass of the block; #\qquad v_0# - speed of the block. #k# - spring constant; #\quad \qquad x# - compression of the spring;

Decrease in KE : #\Delta K = -1/2mv_0^2;# Increase in PE : #\Delta U = + 1/2kx^2#

#\Delta E = 0 \qquad => \Delta U = - \Delta K# #1/2 kx^2 = 1/2 mv_0^2; \qquad x = \sqrt{m/k}v_0 = \sqrt{(9\quad kg)/(6\quad N/m)}(4\quad m/s) = 4.899\quad m#

Parker Abbott · Answer

undefined Using the conservation of mechanical energy, the initial kinetic energy of the object is equal to the potential energy stored in the compressed spring. The initial kinetic energy of the object is $ \frac{1}{2} m v^2 $, where $ m $ is the mass of the object and $ v $ is its initial speed. The potential energy stored in the compressed spring is $ \frac{1}{2} k x^2 $, where $ k $ is the spring constant and $ x $ is the compression of the spring. Setting these equal and solving for $ x $, we find $ x = \sqrt{\frac{m v^2}{k}} $. Plugging in the given values, we get $ x = \sqrt{\frac{(9 \, \text{kg})(4 \, \text{m/s})^2}{6 \, \text{kg/s}^2}} $. Solving this yields $ x \approx 4 \, \text{m} $. Therefore, the spring will compress by approximately 4 meters.