Question

A massless spring with force constant
632 N/m is fastened at its left end to a vertical
wall, as shown below.
The acceleration of gravity is 9.8 m/s^2.
Initially, the 8 kg block and 3 kg block rest
on a horizontal surface with the 8 kg block in
contact with the spring (but not compressing
it) and with the 3 kg block in contact with the
8 kg block. The 8 kg block is then moved to
the left, compressing the spring a distance of
0.2 m, and held in place while the 3 kg block
remains at rest as shown below.
Determine the elastic energy U stored in
the compressed spring.
Answer in units of J.

Answer 1

The elastic energy \( U \) stored in a compressed or stretched spring can be calculated using the formula:

\[
U = \frac{1}{2} k x^2
\]

where:
- \( U \) is the elastic potential energy,
- \( k \) is the spring constant,
- \( x \) is the compression or extension of the spring from its equilibrium position.

Given:
- Spring constant \( k = 632 \, \text{N/m} \)
- Compression distance \( x = 0.2 \, \text{m} \)

Now we can substitute these values into the formula:

\[
U = \frac{1}{2} \times 632 \, \text{N/m} \times (0.2 \, \text{m})^2
\]

First, calculate \( (0.2 \, \text{m})^2 \):

\[
(0.2)^2 = 0.04 \, \text{m}^2
\]

Now substitute back into the equation:

\[
U = \frac{1}{2} \times 632 \times 0.04
\]

Calculating the product:

\[
U = \frac{1}{2} \times 632 \times 0.04 = \frac{1}{2} \times 25.28 = 12.64 \, \text{J}
\]

So, the elastic energy stored in the compressed spring is:

\[
\boxed{12.64 \, \text{J}}
\]