Question

it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additional 0.4 m?

Answer 1

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

$k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m$

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

$E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J$

Therefore, the additional work needed is

$\Delta E=E'-E=810-90=720 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly