Two springs,each with spring constant 20 N/m,are connected in parallel,and the combination is hung vertically from one end.A 10-N weight is then suspended from the other end of the combined spring.How far does the combined spring stretch as it comes to equilibrium?
A)0.50 m
B)1.0 m
C)2.0 m
D)0.25 m

Question

Quizplus · Accepted Answer

When springs are in parallel, their spring constants add up. So, the effective spring constant is 20 N/m + 20 N/m = 40 N/m. Using Hooke's Law (F = kx), where F is the force (10 N), k is the spring constant (40 N/m), and x is the stretch, we solve for x: x = F/k = 10 N / 40 N/m = 0.25 m.

Two Springs,each with Spring Constant 20 N/m,are Connected in Parallel,and