Answer :
Sure, let's break down the solution step by step to find the potential energy stored in a spring that is compressed.
Given Values:
- Compression distance ([tex]$x$[/tex]): [tex]\(0.65 \, \text{m}\)[/tex]
- Mass of the block: [tex]\(25 \, \text{kg}\)[/tex] (not directly needed for this calculation)
- Spring constant ([tex]$k$[/tex]): [tex]\(95 \, \text{N/m}\)[/tex]
Formula:
The potential energy ([tex]\(PE\)[/tex]) stored in a compressed or stretched spring is given by the formula:
[tex]\[ PE = \frac{1}{2} k x^2 \][/tex]
where:
- [tex]\(k\)[/tex] is the spring constant,
- [tex]\(x\)[/tex] is the compression distance.
Calculation:
1. First, square the compression distance:
[tex]\[ x^2 = (0.65)^2 = 0.4225 \, \text{m}^2 \][/tex]
2. Then, multiply this result by the spring constant:
[tex]\[ k \cdot x^2 = 95 \times 0.4225 = 40.1375 \, \text{N} \cdot \text{m} \][/tex]
3. Finally, multiply by [tex]\( \frac{1}{2} \)[/tex] to find the potential energy:
[tex]\[ PE = \frac{1}{2} \times 40.1375 = 20.06875 \, \text{J} \][/tex]
So, the potential energy stored in the spring when it is compressed by [tex]\(0.65 \, \text{m}\)[/tex] is [tex]\(20.06875 \, \text{J}\)[/tex].
Rounding the answer to the nearest whole number, we get:
[tex]\[ PE \approx 20 \, \text{J} \][/tex]
Therefore, the correct option is [tex]\(20 \, J\)[/tex].
Given Values:
- Compression distance ([tex]$x$[/tex]): [tex]\(0.65 \, \text{m}\)[/tex]
- Mass of the block: [tex]\(25 \, \text{kg}\)[/tex] (not directly needed for this calculation)
- Spring constant ([tex]$k$[/tex]): [tex]\(95 \, \text{N/m}\)[/tex]
Formula:
The potential energy ([tex]\(PE\)[/tex]) stored in a compressed or stretched spring is given by the formula:
[tex]\[ PE = \frac{1}{2} k x^2 \][/tex]
where:
- [tex]\(k\)[/tex] is the spring constant,
- [tex]\(x\)[/tex] is the compression distance.
Calculation:
1. First, square the compression distance:
[tex]\[ x^2 = (0.65)^2 = 0.4225 \, \text{m}^2 \][/tex]
2. Then, multiply this result by the spring constant:
[tex]\[ k \cdot x^2 = 95 \times 0.4225 = 40.1375 \, \text{N} \cdot \text{m} \][/tex]
3. Finally, multiply by [tex]\( \frac{1}{2} \)[/tex] to find the potential energy:
[tex]\[ PE = \frac{1}{2} \times 40.1375 = 20.06875 \, \text{J} \][/tex]
So, the potential energy stored in the spring when it is compressed by [tex]\(0.65 \, \text{m}\)[/tex] is [tex]\(20.06875 \, \text{J}\)[/tex].
Rounding the answer to the nearest whole number, we get:
[tex]\[ PE \approx 20 \, \text{J} \][/tex]
Therefore, the correct option is [tex]\(20 \, J\)[/tex].
