Question

"A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 65 N.

a. How much work is done in stretching the spring 5.5 m from its equilibrium​ position?
b. How much work is done in compressing the spring 3.5 m from its equilibrium​ position?
c. Set up the integral that gives the work done in stretching the spring 5.5 m from its equilibrium position. Use increasing limits of integration."

Answer 1

Part a)

`W = 1966.25 J`

Part b)

`W = 796.25 J`

Part c)

`W = \int_(x=0)^(x = 5.5) (130x) dx`

Step-by-step explanation:

Part a)

As we know that 65 N force is required to pull the spring by x = 0.5 m

so we will have

`F = kx`

here we know that

`65 = k(0.5)`

`k = 130 N/m`

now we need to find the work to stretch it by 5.5 m from equilibrium position

So it is given as

`W = (1)/(2)kx^2`

`W = (1)/(2)(130)(5.5^2)`

`W = 1966.25 J`

Part b)

Work done to compress the spring by 3.5 m is given as

`W = (1)/(2)kx^2`

`W = (1)/(2)(130)(3.5^2)`

`W = 796.25 J`

Part c)

Work done by variable force is given as

`W = \int F.dx`

so here we need to stretch it from x = 0 to x = 5.5

so we will have

`F = kx = 130(x)`

now work done is given as

`W = \int_(x=0)^(x = 5.5) (130x) dx`