A ball of mass 540 g hangs from a spring whose stiffness is 155 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?

To determine the relaxed length of the spring when it is detached from the ball and laid on a table, 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.

In this case, the spring's stiffness is given as 155 newtons per meter. This means that for every meter the spring is stretched or compressed, it exerts a force of 155 newtons.

The spring is currently stretched by 15 cm, which is equivalent to 0.15 meters. To find the force exerted by the spring, we multiply the displacement by the spring's stiffness:

Force = Stiffness x Displacement
Force = 155 N/m x 0.15 m
Force = 23.25 N

The force exerted by the spring when the ball is attached is equal to the weight of the ball. We can use this information to find the mass of the ball.

Weight = Mass x Gravity
23.25 N = Mass x 9.8 m/s^2
Mass = 23.25 N / 9.8 m/s^2
Mass = 2.37 kg

We are given that the mass of the ball is 540 g, which is equivalent to 0.54 kg. Since the calculated mass is higher than the given mass, it suggests that there is some additional force acting on the ball to balance the weight.

To find the weight of the ball, we can multiply its mass by the acceleration due to gravity:

Weight = Mass x Gravity
Weight = 0.54 kg x 9.8 m/s^2
Weight = 5.292 N

So, the additional force acting on the ball is:

Additional Force = Calculated Force - Weight
Additional Force = 23.25 N - 5.292 N
Additional Force = 17.958 N

The relaxed length of the spring, when it is detached from the ball and laid on a table, is when there is no force being exerted on it. Since the weight of the ball is acting downwards, the spring will stretch upwards to counterbalance this force.

Therefore, the relaxed length of the spring would be the length when there is no additional force acting on it.

I'm going to assume the diagram gives the tension in the string, because with what you stated you're missing a degree of freedom. Only two equations can be developed (Sum of forces = 0 in both the x and y direction). However, we're missing the tension's magnitude, spring forces magnitude, and spring forces direction.