Question

a spring with a spring constant value of 2500 \\(\frac{\text{n}}{\text{m}}\\) is compressed 32 cm. a 1.5-kg rock is placed on top of it, then the spring is released. approximately how high will the rock rise?\
\\(\bigcirc\\) 9 m\
\\(\bigcirc\\) 17 m\
\\(\bigcirc\\) 27 m\
\\(\bigcirc\\) 85 m

Explanation:

Step1: Convert units

Convert the compression distance from cm to m: \( x = 32\ \text{cm} = 0.32\ \text{m} \).

Step2: Use conservation of energy

The elastic potential energy of the spring is converted to gravitational potential energy of the rock. The formula for elastic potential energy is \( U_s = \frac{1}{2}kx^2 \), and for gravitational potential energy is \( U_g = mgh \). Setting them equal: \( \frac{1}{2}kx^2 = mgh \).

Step3: Solve for height \( h \)

Rearrange the formula: \( h = \frac{kx^2}{2mg} \). Substitute \( k = 2500\ \frac{\text{N}}{\text{m}} \), \( x = 0.32\ \text{m} \), \( m = 1.5\ \text{kg} \), \( g = 9.8\ \text{m/s}^2 \):
\( h = \frac{2500 \times (0.32)^2}{2 \times 1.5 \times 9.8} \).
Calculate numerator: \( 2500 \times 0.1024 = 256 \).
Calculate denominator: \( 2 \times 1.5 \times 9.8 = 29.4 \).
Then \( h = \frac{256}{29.4} \approx 8.71\ \text{m} \), approximately \( 9\ \text{m} \).

Answer:

9 m (the first option)