A spring with a spring constant value of [tex]$2500 \frac{N}{m}$[/tex] is compressed [tex]32 \, \text{cm}[/tex]. A [tex]1.5 \, \text{kg}[/tex] rock is placed on top of it, then the spring is released. Approximately how high will the rock rise?

A. [tex]$9 \, \text{m}$[/tex]

B. [tex]$17 \, \text{m}$[/tex]

C. [tex]$27 \, \text{m}$[/tex]

D. [tex]$85 \, \text{m}$[/tex]



Answer :

Certainly! Let's solve this problem step-by-step.

### Step 1: Understand the Given Values
- Spring constant (k): [tex]\(2500 \frac{N}{m}\)[/tex]
- Compression of the spring (x): [tex]\(32 \, cm = 0.32 \, m\)[/tex] (since 1 cm = 0.01 m)
- Mass of the rock (m): [tex]\(1.5 \, kg\)[/tex]
- Acceleration due to gravity (g): [tex]\(9.8 \frac{m}{s^2}\)[/tex]

### Step 2: Calculate Potential Energy Stored in the Spring
When the spring is compressed, it stores elastic potential energy. The potential energy (PE_spring) stored in a compressed spring is given by:

[tex]\[ PE_{spring} = \frac{1}{2} k x^2 \][/tex]

Using the given values:
[tex]\[ k = 2500 \, \frac{N}{m} \][/tex]
[tex]\[ x = 0.32 \, m \][/tex]

Substitute these values into the formula:

[tex]\[ PE_{spring} = \frac{1}{2} \times 2500 \, \frac{N}{m} \times (0.32 \, m)^2 \][/tex]
[tex]\[ PE_{spring} = \frac{1}{2} \times 2500 \times 0.1024 \][/tex]
[tex]\[ PE_{spring} = 1250 \times 0.1024 \][/tex]
[tex]\[ PE_{spring} = 128 \, J \][/tex]

### Step 3: Convert Spring Potential Energy to Gravitational Potential Energy
When the spring is released, the stored potential energy is converted into gravitational potential energy (PE_gravitational), which will raise the rock to a certain height [tex]\( h \)[/tex]. The formula for gravitational potential energy is:

[tex]\[ PE_{gravitational} = mgh \][/tex]

Where:
[tex]\[ m = 1.5 \, kg \][/tex]
[tex]\[ g = 9.8 \frac{m}{s^2} \][/tex]

We know that the spring potential energy will equal the gravitational potential energy at the maximum height.

[tex]\[ PE_{spring} = PE_{gravitational} \][/tex]
[tex]\[ 128 \, J = 1.5 \, kg \times 9.8 \frac{m}{s^2} \times h \][/tex]

### Step 4: Solve for the Height (h)
Rearrange the equation to solve for [tex]\( h \)[/tex]:

[tex]\[ h = \frac{128 \, J}{1.5 \, kg \times 9.8 \frac{m}{s^2}} \][/tex]
[tex]\[ h = \frac{128}{1.5 \times 9.8} \][/tex]
[tex]\[ h = \frac{128}{14.7} \][/tex]
[tex]\[ h \approx 8.707 \, m \][/tex]

### Final Answer
The rock will rise approximately [tex]\(8.707 \, m\)[/tex]. Hence, the approximately closest answer is:

[tex]\[ 9 \, m \][/tex]

Thus, the correct answer to the question is:
[tex]\[ 9 \, m \][/tex]