A massless spring with force constant

Question

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:

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