Questions about a spring-mass system hanging vertically

[emmalyn1997]

Homework Statement

I would like to see if I did these problems correctly

1) A spring with coefficient k suspends an object of mass m. The object is displaced a distance x from its equilibrium position. The spring hangs vertically from the ceiling. What is the acceleration of the object when it returns to equilibrium? What is the velocity of the mass at its largest displacements?

2) A spring of coefficient 5N/m is hung vertically from the ceiling 10m high. A mass of 5kg is suspended from the spring and displaced 1m upwards. How fast is the mass moving when it is .7m below the equilibrium position?

Relevant Equations

1/2kx^2=1/2mv^2

1) a=0 and v=0

2) 0.7 m/s

 

[BvU]

Homework Statement: I would like to see if I did these problems correctly

1) A spring with coefficient k suspends an object of mass m. The object is displaced a distance x from its equilibrium position. The spring hangs vertically from the ceiling. What is the acceleration of the object when it returns to equilibrium? What is the velocity of the mass at its largest displacements?

2) A spring of coefficient 5N/m is hung vertically from the ceiling 10m high. A mass of 5kg is suspended from the spring and displaced 1m upwards. How fast is the mass moving when it is .7m below the equilibrium position?
Relevant Equations: 1/2kx^2=1/2mv^2

1) a=0 and v=0

2) 0.7 m/s

Hello @emmalyn1997 ,

$\qquad$ !​

I can't tell if you did these problems correctly: you don't show what you did, but only post some numbers
And: what about mgh in your relevant equation ?

I also have a problem with exercise 2: is there a hole in the floor ?

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

$\$

 

[emmalyn1997]

Hello @emmalyn1997 ,

$\qquad$ !​

I can't tell if you did these problems correctly: you don't show what you did, but only post some numbers
And: what about mgh in your relevant equation ?

I also have a problem with exercise 2: is there a hole in the floor ?

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

$\$

For problem 1, I found the acceleration was equal to zero because at equilibrium the net force would be equal to zero, and plugging it into F=ma to be 0=ma, causes the acceleration to be zero. I found velocity to be equal to zero because the energy is all potential energy, therefore no kinetic energy meaning velocity would be zero.
As for problem 2, I am not exactly sure about your question, that was just the problem as worded on the study guide. I originally used the equation 1/2kx^2=mgh, and then I used the conservation of mechanical energy to relate potential energy to kinetic energy of the mass, ending up with the equation 1/2kx^2=1/2mv^2. And then solving for velocity v=(sqrt k/m) * x. Getting the solution 0.7m/s. I am not quite sure what other equation to use for this problem so if you have any other ideas that would be very helpful.

 

[PeroK]

For problem 1, I found the acceleration was equal to zero because at equilibrium the net force would be equal to zero, and plugging it into F=ma to be 0=ma, causes the acceleration to be zero. I found velocity to be equal to zero because the energy is all potential energy, therefore no kinetic energy meaning velocity would be zero.

That's correct.

As for problem 2, I am not exactly sure about your question, that was just the problem as worded on the study guide. I originally used the equation 1/2kx^2=mgh, and then I used the conservation of mechanical energy to relate potential energy to kinetic energy of the mass, ending up with the equation 1/2kx^2=1/2mv^2. And then solving for velocity v=(sqrt k/m) * x. Getting the solution 0.7m/s. I am not quite sure what other equation to use for this problem so if you have any other ideas that would be very helpful.

What is $x$ here?

 

[haruspex]

is there a hole in the floor ?

Moreover, we are not told the relaxed length of the spring.
@emmalyn1997 , this may be a trick question. Draw a diagram of the equilibrium position.

 

[emmalyn1997]

That's correct.

What is $x$ here?

Sorry! X is the displacement so in this situation it would be 0.7 because it is 0.7m below equilibrium.

 

[PeroK]

Sorry! X is the displacement so in this situation it would be 0.7 because it is 0.7m below equilibrium.

What was the relevance of displacing the mass by $1m$?

What do you get if the mass is $1m$ below the equilibrium point. I.e. if $x = 1m$?

 

[PeroK]

Here's an interesting numerical observation:
$$1^2 - 0.7^2 = 0.51 \approx 0.7^2$$Isn't that funny?