Question

1. a pop - up toy has a mass of 0.020 kilogram and a spring constant of 150 newtons per meter. a force is applied to the toy to compress the spring 0.050 meter.

a. calculate the potential energy stored in the compressed spring.

Explanation:

Step1: Recall the formula for elastic potential energy

The formula for the elastic potential energy \( U \) stored in a spring is \( U = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position (compression or extension).

Step2: Identify the given values

We are given:

• Spring constant \( k = 150 \, \text{N/m} \)
• Compression \( x = 0.050 \, \text{m} \)

Step3: Substitute the values into the formula

Substitute \( k = 150 \, \text{N/m} \) and \( x = 0.050 \, \text{m} \) into the formula \( U = \frac{1}{2}kx^2 \):

\[

$$\begin{align*} U &= \frac{1}{2} \times 150 \, \text{N/m} \times (0.050 \, \text{m})^2 \\ &= \frac{1}{2} \times 150 \times 0.0025 \\ &= 75 \times 0.0025 \\ &= 0.1875 \, \text{J} \end{align*}$$

\]

Answer:

The potential energy stored in the compressed spring is \( 0.1875 \, \text{joules} \) (or approximately \( 0.19 \, \text{J} \) if rounded to two significant figures).