What is the potential energy of a spring that is compressed [tex]$0.65 \, \text{m}$[/tex] by a [tex]25 \, \text{kg}$[/tex] block if the spring constant is [tex]$95 \, \text{N/m}$[/tex]?

A. 1.6 J
B. 7.9 J
C. 15 J
D. 20 J



Answer :

To find the potential energy stored in a compressed spring, we can use the formula for elastic potential energy:

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

where [tex]\( PE \)[/tex] is the potential energy, [tex]\( k \)[/tex] is the spring constant, and [tex]\( x \)[/tex] is the displacement (compression) of the spring.

Given the values:
- Displacement, [tex]\( x = 0.65 \)[/tex] meters
- Spring constant, [tex]\( k = 95 \)[/tex] N/m

We can now substitute these values into the formula:

[tex]\[ PE = \frac{1}{2} \times 95 \times (0.65)^2 \][/tex]

First, calculate [tex]\( x^2 \)[/tex]:

[tex]\[ (0.65)^2 = 0.4225 \][/tex]

Next, multiply this value by the spring constant [tex]\( k \)[/tex]:

[tex]\[ 95 \times 0.4225 = 40.1375 \][/tex]

Then, multiply by [tex]\( \frac{1}{2} \)[/tex]:

[tex]\[ \frac{1}{2} \times 40.1375 = 20.06875 \][/tex]

Therefore, the potential energy stored in the spring is approximately 20 J.

So, the correct answer is:

[tex]\[ \boxed{20 \text{ J}} \][/tex]