Question

It takes a force of 28,848 lb to compress a coil spring assembly from its free height of 8 inches to its fully compressed height of 4 inches. a. What is the​ assembly's force​ constant? b. How much work does it take to compress the assembly the first half​ inch? the second half​ inch? Answer to the nearest​ in-lb.

Answer 1

k = 7212

W =2705 in-lb

Step-by-step explanation:

force (F) = 28,848 lb

we know,

a) F = k x ..........................(1)

x = 8 - 4 = 4 inch

from equation (1)

28,848 = k × 4

k = 7212 (assembly force constant)

b) we know

Force at x distance F(x) = 7212x

work done = force × displacement

assume displacement of small distance dx

work done = 7212x dx

for first half inch

`W = \int\limits^(0.5)_0 7212x\ dx\\W = 3606\ [x^2]^(0.5)_0\\W=901.5\ inch-pound`

hence for second half inch we just multiply first half with 3

W = 3×901.5

W = 2704.5 ≈ 2705 in-lb