The spring of a spring balance is 5.0in. long when there is no weight on the balance, and it is 8.4in. long with 4.0 lb hung from the balance. How much work is done in stretching it from 5.0in. to a length of 12.8in.?

work is done in stretching it from 5.0in. to a length of 12.8in is 36 lb.In

Step-by-step explanation:

We know that

f(x) = kx

given that;

Natural length L = 5

x = 8.4 - 5 = 3.4

force F = 4.0 lb

f(x) = kx

we substitute

4 = k × 3.4

k = 4 / 3.4

k = 1.1764 = 20/17

Now work done from natural length to 12.8

x = 12.8 - 5 = 7.8

work done = ₀∫^7.8  f(x) dx

= ₀∫^7.8  kx dx

we substitute value of k

= ₀∫^7.8  (20/17)x dx

= 20/17 [ x²/2 ]

= 20/17 [ (7.8)²/2 - 0 ]

= 1.1764 × 30.42

= 35.786 ≈ 36 lb.In

Therefore work is done in stretching it from 5.0in. to a length of 12.8in is 36 lb.In

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)