Question

base your answers to questions 71 through 75 on the information and diagram below and on your knowledge of physics. a spring with a spring constant of 2600 newtons per meter is compressed 0.10 meter from its unstretched position. the spring is released, propelling a 3.0 - kilogram block along a horizontal, frictionless surface. this block then collides with a stationary 1.0 - kilogram block. the blocks remain joined and move together as shown in the diagram below. assuming all of the springs energy is transferred to the 3.0 - kilogram block, calculate the speed, v1, of the 3.0 - kilogram block immediately after it is propelled by the spring.

Explanation:

Step1: Calculate initial spring - potential energy

The formula for spring - potential energy is $U_s=\frac{1}{2}kx^{2}$, where $k = 2600\ N/m$ and $x=0.10\ m$.
$U_s=\frac{1}{2}\times2600\times(0.10)^{2}$
$U_s = 13\ J$

Step2: Apply conservation of energy to find speed of 3.0 - kg block

Just after the spring is released and before the collision, all of the spring's potential energy is converted into the kinetic energy of the 3.0 - kg block. The formula for kinetic energy is $K=\frac{1}{2}mv^{2}$.
Since $U_s = K$ and $m = 3.0\ kg$, we have $\frac{1}{2}mv^{2}=13\ J$.
$v=\sqrt{\frac{2\times13}{3}}$
$v=\sqrt{\frac{26}{3}}\approx2.94\ m/s$

Answer:

$2.94\ m/s$