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?

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.

To find the distance the combined spring stretches as it comes to equilibrium, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

First, let's find the effective spring constant (k) for the parallel combination of the two springs. When springs are connected in parallel, the effective spring constant is the sum of the individual spring constants.

Given:
Spring constant of each spring (k) = 20 N/m
Weight suspended (W) = 10 N

The force exerted by the combined spring (F) is equal to the weight suspended. So, F = W.

Hence, we can write the equation as:
F = k * x

where F is the force, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

Since we have two springs in parallel, the effective spring constant (k) will be the sum of the individual spring constants:
k = k1 + k2
= 20 N/m + 20 N/m
= 40 N/m

Now, we can rearrange the equation to solve for x:
x = F / k

Substituting the known values:
x = 10 N / 40 N/m
x = 0.25 m

Therefore, the combined spring stretches 0.25 meters as it comes to equilibrium.

To find the distance that the combined spring stretches when it comes to equilibrium, we can use Hooke's Law which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.

In this case, we have two springs connected in parallel, and the force exerted by each spring is given by F = k * x, where F is the force, k is the spring constant, and x is the displacement.

Since the two springs are connected in parallel, they will experience the same force when the system reaches equilibrium. Therefore, the total force exerted on the combined spring is equal to the sum of the forces exerted by the individual springs.

Let's assume the displacement of the combined spring is x_total. The force exerted by each individual spring is given by F1 = k1 * x_total and F2 = k2 * x_total, where k1 and k2 are the spring constants of the individual springs.

The total force exerted on the combined spring is F_total = F1 + F2 = k1 * x_total + k2 * x_total.

According to Newton's second law, the force exerted by the weight is equal to the total force exerted on the combined spring. Therefore, we have:

F_weight = F_total

Substituting the values into the equation, we get:

10 N = (k1 + k2) * x_total

Given that k1 = k2 = 20 N/m, we can simplify the equation:

10 N = (20 N/m + 20 N/m) * x_total

10 N = (40 N/m) * x_total

Dividing both sides of the equation by 40 N/m, we get:

0.25 m = x_total

Therefore, the combined spring stretches 0.25 meters (or 25 centimeters) when it comes to equilibrium.