Why Isn't the Right Side Negative in a Vertical Spring-Mass System Equation?

[Ark236]

Homework Statement

Hi everyone,

The problem has two parts. The first is to determine the equilibrium position of a mass attached to a spring. The second is to determine the equation of motion of the system, assuming that the block is pulled 1 cm down from its equilibrium position.

Relevant Equations

I choose the downward direction as positive. For the first part and using the FBD:

mg - k y_{0} = 0

Then the equilibrium position is y_{0} = mg/k.

For the second part, we have that:

mg -k y = m d^2 y/dt^2

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards.

thanks

image was obtained from here

Thanks.

 

[Hill]

Homework Statement: Hi everyone,

The problem has two parts. The first is to determine the equilibrium position of a mass attached to a spring. The second is to determine the equation of motion of the system, assuming that the block is pulled 1 cm down from its equilibrium position.
Relevant Equations: I choose the downward direction as positive. For the first part and using the FBD:

mg - k y_{0} = 0

Then the equilibrium position is y_{0} = mg/k.

For the second part, we have that:

mg -k y = m d^2 y/dt^2

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards.

thanks

image was obtained from here

Thanks.View attachment 334759

When $y>y_0, mg-ky<0$, and then the right side, $m \frac {d^2y} {dt^2}$ is negative.

 

[nasu]

Both sides are negative at that point. The net force points upwards and the body accelerates upwards. The equation shows that thenet force and the acceleration have the same sign. They are either both positive or both negative. If you put a minus sign in the equation itself, it would mean that the acceleration is in direction opposite to the net force. This would contradict Newton's second law, wouldn't?

 

[Steve4Physics]

mg -k y = m d^2 y/dt^2

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards.

When the mass is below it's equilibrium position, which is bigger: mg or ky?