Question

A block of mass 2.00 kg is placed against a horizontal spring of constant k = 855 N/m and pushed so the spring compresses by 0.0400 m.
(a)
What is the elastic potential energy of the block-spring system (in J)?
J
(b)
If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring.
m/s

Answer 1

To solve the given problem, we will use the formulas for elastic potential energy and the relationship between potential energy and kinetic energy.

### (a) Elastic Potential Energy

The elastic potential energy \( U \) stored in a compressed or stretched spring is given by the formula:

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

where:
- \( k \) is the spring constant,
- \( x \) is the compression or extension of the spring.

Given:
- \( k = 855 \, \text{N/m} \)
- \( x = 0.0400 \, \text{m} \)

Substituting these values into the formula:

\[
U = \frac{1}{2} \times 855 \, \text{N/m} \times (0.0400 \, \text{m})^2
\]

Calculating this:

\[
U = \frac{1}{2} \times 855 \times 0.0016 \, \text{m}^2 = \frac{1}{2} \times 1.368 \, \text{J} \approx 0.684 \, \text{J}
\]

Thus, the elastic potential energy of the block-spring system is:

\[
\boxed{0.684 \, \text{J}}
\]

### (b) Block's Speed After Leaving the Spring

When the block is released and the surface is frictionless, all the elastic potential energy will convert into kinetic energy. The kinetic energy \( K \) is given by:

\[
K = \frac{1}{2} mv^2
\]

Setting the kinetic energy equal to the potential energy:

\[
\frac{1}{2} mv^2 = U
\]

Substituting the values we know:

- \( m = 2.00 \, \text{kg} \)
- \( U = 0.684 \, \text{J} \)

Thus,

\[
\frac{1}{2} \times 2.00 \, \text{kg} \times v^2 = 0.684 \, \text{J}
\]

This simplifies to:

\[
1.00 \, \text{kg} \times v^2 = 0.684 \, \text{J}
\]

Solving for \( v^2 \):

\[
v^2 = 0.684 \, \text{m}^2/\text{s}^2
\]

Taking the square root to find \( v \):

\[
v = \sqrt{0.684} \approx 0.826 \, \text{m/s}
\]

Thus, the speed of the block after leaving the spring is:

\[
\boxed{0.826 \, \text{m/s}}
\]