Answer:
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
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
Learn more here: brainly.com/question/2918375
