Question

base your answers to questions 71 through 73 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. determine the total amount of elastic potential energy stored in the spring when the spring is compressed 0.10 meter.

Explanation:

Step1: Identify the formula for elastic - potential energy

The formula for elastic - potential energy stored in a spring is $U = \frac{1}{2}kx^{2}$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium (unstretched) position.

Step2: Substitute the given values into the formula

We are given that $k = 2600\ N/m$ and $x=0.10\ m$.
$U=\frac{1}{2}(2600\ N/m)(0.10\ m)^{2}$

Step3: Calculate the value of elastic - potential energy

First, calculate $(0.10\ m)^{2}=0.01\ m^{2}$. Then, $\frac{1}{2}(2600\ N/m)(0.01\ m^{2})$.
$\frac{1}{2}\times2600\times0.01 = 13\ J$

Answer:

$13\ J$