How much work is required to stretch an ideal spring of spring constant (force constant) 40
N/m from x = 0.20 m to x = 0.25 m if the unstretched position is at x = 0.00 m?
A) 0.45 J
B) 0.050 J
C) 0.80 J
D) 1.3J

The definition of work and Hooke's law allows to find the correct answer for the work in the spring is:

    A)  The work is:   0.45 J

Work is defined by the dot product between the force and the displacement.

         dW = F . dx

Where bold indicates vectors, f is force and d is displacement.

Hooke's law states that the restoring force in a spring is proportional to its displacement from its equilibrium position.

         F = - k x

We substitute

         dW = - k x dx

The displacement is in the opposite direction of the force, dx = -dx

We integrate and evaluate from work 0 to W for the displacement of the lower limit x₀ or to the upper limit x.

        W = ½ k (x²- x₀²)

They indicate that the spring constant is k = 40 N / m and the displacement of from x₀ = 0.20 m to x = 0.25 m

Let's calculate

         W = ½ 40 (0.25² - 0.20²)

         W = 0.45 J

In conclusion using the definition of work and Hooke's law we can find the correct answer for the work in the spring is:

    A) The  work is:  0.45 J

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0.05J, option B)

Explanation:

Work=k(dx)²/2, where dx is the actual deformation.

dx=x2-x1=0.25-0.2=0.05m

(dx)²=0.05²=0.0025

Work=40×0.0025/2=20×0.0025=2×0.025=0.05J

does anyone know or what?