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?

1 Answer
Mar 9, 2016

Answer:

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

Explanation:

Mechanical Energy Conservation: When the forces acting on a system are purely conservative, its mechanical energy remains conserved.

#\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 case when the kinetic energy of the block decreases, the potential energy in the spring increases by the same amount to keep the total mechanical energy of the system a constant.

#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#