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.

Question

(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:

$$\boxed{{\displaystyle 0.684,\text{J}}}$$

(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:

$$K=\frac{1}{2}m{v}^2$$

Setting the kinetic energy equal to the potential energy:

$$\frac{1}{2}m{v}^2=U$$

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:

$$\boxed{{\displaystyle 0.826,\text{m/s}}}$$