if a force of 50 N stretches a spring 250 mm, what is the elastic potential energy of the spring when it is stretched by 200 mm?

To calculate the elastic potential energy of a spring, you need to use Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) of the spring from its equilibrium position.

Hooke's Law equation: F = k * x

Where:
F = Force applied to the spring (in Newtons)
k = Spring constant (in N/m)
x = Displacement of the spring (in meters)

From the given information, we can determine the spring constant (k) using Hooke's Law equation. Rearranging the equation, we get:

k = F / x

Plugging in the values: F = 50 N and x = 250 mm = 0.25 m, we can calculate the spring constant:

k = 50 N / 0.25 m
k = 200 N/m

Now that we know the spring constant (k), we can calculate the elastic potential energy using the formula:

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

To find the elastic potential energy when the spring is stretched by 200 mm = 0.2 m, plug in the values:

U = (1/2) * 200 N/m * (0.2 m)^2
U = 4 J (Joules)

So, the elastic potential energy of the spring when stretched by 200 mm is 4 Joules.