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.

k = 7212

W =2705 in-lb

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

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

W = 3×901.5

W = 2704.5 ≈ 2705 in-lb

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

iron would move it is magnetic there is a mag field around the current in the wire