A ball of mass 620 g hangs from a spring whose stiffness is 195 newtons per meter. A string is attached to the ball and you are pulling the string to the right, so that the ball hangs motionless, as shown in the figure. In this situation the spring is stretched, and its length is 15 cm. What would be the relaxed length of the spring, if it were detached from the ball and laid on a table?

Why did the spring go on vacation? Because it needed some "me time" to relax! All jokes aside, to find the relaxed length of the spring, 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, since the ball is motionless, the force exerted by the spring is equal to the weight of the ball.

First, let's convert the mass of the ball from grams to kilograms:
Mass = 620 g = 0.62 kg

Now, let's find the force exerted by the spring:
Force = mass x gravity
Force = 0.62 kg x 9.8 m/s^2
Force = 6.076 N

According to Hooke's Law, the force exerted by the spring is also equal to the stiffness of the spring multiplied by the displacement. In this case, the displacement is the difference between the stretched length of the spring and the relaxed length of the spring. Let's assume the relaxed length of the spring is x meters.

6.076 N = 195 N/m x (0.15 m - x)

Simplifying the equation:

0.15 m - x = 6.076 N / 195 N/m
0.15 m - x = 0.0312 m

Now, let's solve for x:

x = 0.15 m - 0.0312 m
x = 0.1188 m

So, the relaxed length of the spring would be approximately 0.1188 meters.

To find the relaxed length of the spring, we need to understand the concept of the equilibrium position. In an equilibrium position, the net force on an object is zero. In this case, the ball is hanging motionless, which means the net force acting on it is zero.

Using Hooke's Law, we can establish an equation to find the equilibrium position of the spring, where the force exerted by the spring is equal to the weight of the ball. The equation is:

F_spring = k * Δl

Where:
F_spring is the force exerted by the spring
k is the stiffness of the spring
Δl is the change in length of the spring

The weight of the ball is given by:

F_gravity = m * g

Where:
m is the mass of the ball
g is the acceleration due to gravity

Since the ball is hanging motionless, we can set up the following equation:

F_spring = F_gravity

k * Δl = m * g

Now, let's plug in the given values:

k = 195 N/m (stiffness of the spring)
Δl = 15 cm = 0.15 m (change in length of the spring)
m = 0.620 kg (mass of the ball)
g ≈ 9.8 m/s^2 (acceleration due to gravity)

Rearranging the equation, we can solve for the change in length of the spring:

Δl = (m * g) / k

Substituting the given values:

Δl = (0.620 kg * 9.8 m/s^2) / 195 N/m
Δl ≈ 0.031 m

The relaxed length of the spring, if it were detached from the ball and laid on a table, would be the sum of the stretched length and the change in length of the spring:

Relaxed Length = Stretched Length + Δl

Given that the stretched length is 15 cm or 0.15 m, we can calculate the relaxed length:

Relaxed Length = 0.15 m + 0.031 m
Relaxed Length ≈ 0.18 m

Therefore, the relaxed length of the spring would be approximately 0.18 meters.