A block of mass 3.80 kg is placed against a horizontal spring of constant k = 815 N/m and pushed so the spring compresses by 0.0750 m.
(a) What is the elastic potential energy of the block-spring system (in 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.

Let's solve this step-by-step.

### Part (a): Elastic Potential Energy of the Block-Spring System

The elastic potential energy (\( U \)) stored in a compressed or stretched spring is given by:
\[
U = \frac{1}{2} k x^2
\]
where:
- \( k \) is the spring constant (815 N/m),
- \( x \) is the compression or extension of the spring (0.0750 m).

Substituting the values:
\[
U = \frac{1}{2} \times 815 \, \text{N/m} \times (0.0750 \, \text{m})^2
\]

Calculating this:
\[
U = \frac{1}{2} \times 815 \times 0.005625 = 2.2903 \, \text{J}
\]

So, the elastic potential energy of the system is approximately:
\[
U \approx 2.29 \, \text{J}
\]

### Part (b): Speed of the Block After Leaving the Spring

Since the surface is frictionless, the elastic potential energy of the spring will convert entirely into the kinetic energy of the block once it leaves the spring.

The kinetic energy (\( K \)) of the block is given by:
\[
K = \frac{1}{2} m v^2
\]
where:
- \( m \) is the mass of the block (3.80 kg),
- \( v \) is the speed of the block.

Since all the potential energy is converted to kinetic energy:
\[
U = K
\]
Thus,
\[
2.29 = \frac{1}{2} \times 3.80 \times v^2
\]

Rearrange to solve for \( v \):
\[
v^2 = \frac{2 \times 2.29}{3.80}
\]
\[
v^2 = \frac{4.58}{3.80} \approx 1.2053
\]
\[
v \approx \sqrt{1.2053} \approx 1.10 \, \text{m/s}
\]

So, the speed of the block after leaving the spring is approximately:
\[
v \approx 1.10 \, \text{m/s}
\]

### Summary:
- (a) Elastic potential energy: \( 2.29 \, \text{J} \)
- (b) Speed of the block: \( 1.10 \, \text{m/s} \)

Would you like more details on any part of the solution or have any questions?

---

Here are five related questions for further understanding:
1. How would the speed change if the mass of the block were doubled?
2. What would happen if there was friction on the surface?
3. How does the spring constant affect the energy stored in the spring?
4. What is the relationship between kinetic and potential energy in this scenario?
5. How would the result differ if the block was initially compressed on an inclined plane?

**Tip:** For springs, the energy is proportional to the square of the compression or extension, so doubling the compression increases the energy fourfold.

A block of mass 3.80 kg is placed against a horizontal spring of constant k = 815 N/m and pushed so the spring compresses by 0.0750 m. (a) What is the elastic potential energy of the block-spring system (in 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.

Solve a physics problem involving a 3.80 kg block compressed against a spring with k = 815 N/m. Learn to calculate the elastic potential energy and the block's speed on a frictionless surface. This problem covers key physics concepts, including energy conservation, elastic potential energy, and kinetic energy.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation