Mass on a vertical string - Conservation of Energy Problem

In summary, the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a), and then it is pushed upward with an initial speed of vi = 2 m/s. The maximum speed of the block is 4.1 m/s.
  • #1
Bryon
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0

Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-08-GPE-ME/mass_vertical_spring/7.gif

A spring with spring constant k = 45 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 1.5 kg is hung gently on the end of the spring.

a) How far does the spring stretch? Answer = 0.327m

Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 2 m/s.

b) What is the maximum speed of the block? <---this one I am not sure of.

Homework Equations



0.5*m*v^2 = Ke
0.5*k*x^2 = K(spring)
U + K = Uo + Ko

The Attempt at a Solution



0.5*m*v(final)^2 = 0.5*m*v(initial)^2 + 0.5*k*x^2

for v I got 4.1 m/s. But the answer is incorrect. Any ideas where I went wrong?
 
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  • #2
Your method for (b) is correct. At first I did what you did and got the same numbers as you. Then I reread the problem carefully. It is not very clear, but when it says "the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a)", it means the spring is stretched from the unstretched position by twice the amount not from the new equilibrium by twice the amount. Try it.
 
  • #3
Ah, ok so what I want to find is the length (L) of the relaxed spring with no force applied to it?
 
  • #4
Bryon said:
Ah, ok so what I want to find is the length (L) of the relaxed spring with no force applied to it?
You don't need to find that, in fact you can't find that. All I'm saying is that the x in your last equation should be 0.327 m not 2*0.327 m. It represents the displacement from equilibrium when the mass is hanging and that is 0.327 m.
 
  • #5
Ah, ok I had to read that a few times to understand it. That question is rather confusing, but I think i now see why its not twice the distance. Again, thanks for the help I truly appreciate the guidance!
 

Related to Mass on a vertical string - Conservation of Energy Problem

1. What is the concept of conservation of energy in the context of a mass on a vertical string?

The concept of conservation of energy states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another. In the case of a mass on a vertical string, the potential energy of the mass is converted into kinetic energy as it moves down the string, and vice versa as it moves up the string. The total energy (potential + kinetic) remains constant throughout the motion, demonstrating the principle of conservation of energy.

2. How do you calculate the potential energy of the mass on a vertical string?

The potential energy of the mass is calculated using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above a reference point. In the case of a mass on a vertical string, the reference point is usually the ground or the bottom of the string.

3. Can the mass ever reach a point where it has no potential energy?

No, the mass will always have some potential energy as long as it is above the reference point. Even if the mass is at the top of the string and has no kinetic energy, it will still have potential energy due to its position above the reference point.

4. How does the mass's speed change as it moves up and down the string?

The mass's speed changes based on its position on the string. As it moves down the string, its speed increases due to the conversion of potential energy into kinetic energy. As it moves up the string, its speed decreases as kinetic energy is converted back into potential energy. At the top of the string, the mass's speed is momentarily zero before it begins to move back down.

5. What are the factors that affect the conservation of energy in this problem?

The conservation of energy in this problem is affected by several factors: the mass of the object, the height of the object above the reference point, the acceleration due to gravity, and any external forces acting on the object (such as friction or air resistance). Any changes in these factors can affect the total energy of the system, but the principle of conservation of energy will still hold true.

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