an object is hung from a vertical spring, extending it by 24cm.if the spring constant is 35 N/m, what is the potential energy of the stretched spring ?

The answer I got using the equation is 1.008J but O am not sure of it

To calculate the potential energy of a stretched spring, you can use the following formula:

Potential Energy (PE) = (1/2) * k * (x^2)

Where:
PE is the potential energy of the spring,
k is the spring constant,
x is the displacement or extension of the spring.

In this case, the given values are:
k = 35 N/m (spring constant)
x = 24 cm (extension of the spring)

However, it's important to note that all the values in the equation should be in SI units. So, let's convert the extension from centimeters to meters.

24 cm = 24/100 = 0.24 m

Now, we can substitute the values into the formula and solve for the potential energy.

PE = (1/2) * 35 N/m * (0.24 m)^2

Simplifying the equation:

PE = (1/2) * 35 N/m * 0.0576 m^2

PE = 0.5 * 35 N/m * 0.0576 m^2

PE = 0.5 * 35 N/m * 0.0576 m^2

PE = 0.5 * 2.52 N m

PE = 1.26 N m

Therefore, the potential energy of the stretched spring is 1.26 N m.

To find the potential energy of a stretched spring, you can use the formula:

Potential Energy (PE) = (1/2) * k * x^2

Where:
- PE is the potential energy of the spring.
- k is the spring constant.
- x is the displacement or extension of the spring.

Given:
The spring constant, k = 35 N/m.
The extension of the spring, x = 24 cm = 0.24 m.

Now, let's substitute these values into the formula to find the potential energy of the stretched spring:

PE = (1/2) * k * x^2
= (1/2) * 35 N/m * (0.24 m)^2
= (1/2) * 35 N/m * 0.0576 m^2
= 0.5 * 35 N/m * 0.0576 m^2
≈ 0.5 * 35 N/m * 0.0576 m^2
≈ 0.5 * 2.016 N.m
≈ 1.008 N.m

Therefore, the potential energy of the stretched spring is approximately 1.008 N.m.

pe=1/2 k x^2 x=.24m