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?

Answer 1

When two springs are connected in parallel, their effective spring constant is the sum of their individual spring constants. In this case, the spring constants are the same, so their combined spring constant is:

k_total = k1 + k2 = 20 + 20 = 40 N/m

Now we can use Hooke's Law to find the stretch in the spring:

F = k * x

where F is the force applied, k is the combined spring constant, and x is the stretch in the spring. We can solve for x:

x = F / k = 10 N / 40 N/m = 0.25 m

So the combined spring stretches by 0.25 meters when the 10-N weight is suspended from it.