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

Answer:

Part a)

$W = 1966.25 J$

Part b)

$W = 796.25 J$

Part c)

$W = \int_{x=0}^{x = 5.5} (130x) dx$

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

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

$W = \frac{1}{2}(130)(5.5^2)$

$W = 1966.25 J$

Part b)

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

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

$W = \frac{1}{2}(130)(3.5^2)$

$W = 796.25 J$

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

$W = \int_{x=0}^{x = 5.5} (130x) dx$