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### Course: Physics archive>Unit 5

Lesson 2: Springs and Hooke's law

# Vertical springs and energy conservation

In this video, David explains two different strategies to deal with vertical springs and compares them with those used for horizontal springs. Created by David SantoPietro.

## Want to join the conversation?

• In his equation of summation of the forces acting during the equilibrium point. why isn't the force exerted by the spring -kx?
• great question . you can write in this way : -kx + mg = 0 or kx - mg = 0 . It's same . the negative sign is used to make the spring force exerted in opposite direction ( gravitational force mg vs spring force -kx ) I hope you got it. ;)
• What happens if you attach a mass to two vertical springs with different spring constants?
• it depends how you attach the springs... Parallel or series.
• Suppose instead of a spring it is a sling shot ( same idea) but instead of a vertical launch the launch is at some angle. If I wanted to work through conservation of energy issues to solve for the maximum height obtained for say the stone launched by the sling shot their would be a "horizontal" component of energy ( kinetic energy) to consider. That doesn't disappear.
• Nice thinking.

Think, therefore, where does the energy go?

Seems to me like you have kinetic energy horizontally (which remains constant

and increasing gravitational potential energy vertically

what you think?
• What happens if the string itself is being raised with a constant acceleration?.
• When we let go of the spring, is the force still there? If yes, what is it equal to?
If there was a block attached to the spring, would the spring exert a tension force to the block?
• It is equal to the mass times the instantaneous acceleration of the block. In case there is no block attached, then the mass will be equal to that of the spring itself which shall accelerate (but different points will accelerate differently and shall have different tensions, a very complicated case). Much simpler to understand if we attach a block to a theoretically massless spring.
• So the kinetic energy of the mass is equal to the spring energy when we pull/push the mass and the spring pushes/pulls is back and when the mass is at equilibrium position?
(1 vote)
• The kinetic energy of the spring is equal to its elastic potential energy, i.e. 1/2mv^2 = 1/2kx^2 when the spring is stretched some distance x from the equilibrium point and when its mass also has some velocity, v, with which it is moving. This occurs somewhere in between the equilibrium point and the extreme point (extreme point is when x=amplitude, A).
At the equilibrium, the spring is not stretched any distance away from the equilibrium, i.e. x=0 and thus the mass moves with maximum velocity (as the total energy = kinetic energy + elastic potential energy, and this is conserved). Therefore, at equilibrium, total energy = elastic potential energy.
At the extreme point, the spring is stretched to its max ability as x=A, and the velocity is zero as this is the point just before the velocity is about to change its direction. Therefore, at this point, the total energy = kinetic energy.
I hope that answered your question :)
• i have to find out if the mechanical energy is conserved in an oscillating spring mass apparatus (basically oscillating a mass hanging from a spring. What should be my initial and final mechanical energies? I'm thinking my initial mechanical energy will be when I pull the mass, and my final energy will be when the mass comes to rest after oscillation. Is that right? Can I find out if the energy is conserved that way?
• If it comes to rest after having been moving, was mechanical energy conserved?
• if i stretched a spring and let it go (assuming no energy is lost to heat ,sound...)would it go back and forth forever?
(1 vote)
• Wouldn't gravity cause it to stay down after a long period of time and does't air resistance play a part that would eventually slow down the object enough to bring it to a stop. Even though air resistance barely constitutes to a substantial force it should still play a part in bringing it to a stop
• How do I solve a problem that pertains to a spring sitting on a horizontal plane when a 2kg mass is placed on top of it? I'm given that K= 100N/m and that the spring is compressed .15m, and I need to find the potential energy stored in the spring as well as the maximum height the object ascends.
(1 vote)
• the spring is vertical right?

the mass is sitting on top the spring right? Is the mass causing it to compress? or is the spring compressed After the mass has been placed on it?

there is an equation for energy contained in a spring... you know it?

can you say why the object would ascend?
• So technically, if the block were to just be attached to the spring with someone holding the spring at its original length, and then they let go... The spring would then stretch to a maximum length, oscillate, and come to a new equilibrium position right?
• That sounds right. When the mass is attached to the spring, even though it is being held at the original length, it is still displaced from the equilibrium position, which in this case, is when the force of the spring is balanced out by the force of gravity on the mass (kx = mgh)
(1 vote)