Energy Problem involving Spring

In summary, a 1.0 kg block is attached to two rigid supports by springs A and B. A force of 10.0 N stretches spring A by 0.25 m while a force of 2.5 N extends spring B by the same amount. The block is initially at rest between unstretched springs and is then pushed to the side a distance of 0.50 m, compressing one spring and extending another. The total work done to move the block is 6.25 J. The velocity of the block when released through its original equilibrium position can be determined using the total energy of the system and the potential energy stored in the springs. The spring constant of a single spring that would duplicate A and B
  • #1
testme
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Homework Statement


A block of mass 1.0 kg, at rest on a horizontale table, is attached to two rigid supports by springs A and B. A force of 10.0 N strecthes spring A alone by 0.25 m while a force of 2.5 N extends spring B alone by the same amount. Initially the block is at rest between unstretched springs; then it is pushed to the side a distance of 0.50 m, compressing one spring and extending another.

a) What is the total work done to move the block. (The block is held at rest.)
b) If the block is released with what velocity does it move through its original equilibrium position?
c) What would be the spring constant of a single spring that would duplicate A and B?

mass of block = 1.0 kg
Force to stretch spring A by 0.25 m = 10.0 N
Force to stretch spring B by 0.25 m = 2.5 N
side distance pushed = .5 m

Homework Equations


W = FΔd
Ee = 1/2kx^2
Ek = 1/2mv^2

The Attempt at a Solution


a) W = Fd
= (10 + 2.5)(0.5)
= 6.25 J

b) Not sure how to do this I think it has something to do with the energy but not sure exactly what it's asking for. From what I understood it'd be something like

1/2(k1)x^2 + 1/2(k2)x^2 + 1/2mv^2 = 1/2(k1)x^2 + 1/2(k2)x^2
 
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  • #2
testme said:
a) W = Fd
= (10 + 2.5)(0.5)
= 6.25 J
The force produced by a spring does not remain constant with distance. So you'll either have to integrate F*dx or use the 'canned' expression for spring potential energy.

It happens to be a fluke of the numbers for this problem that your 'method' above gives the same result as the correct methods.
b) Not sure how to do this I think it has something to do with the energy but not sure exactly what it's asking for. From what I understood it'd be something like

1/2(k1)x^2 + 1/2(k2)x^2 + 1/2mv^2 = 1/2(k1)x^2 + 1/2(k2)x^2
Something like that :smile:

What's the total energy that system holds? Give it a name (say, E).

When the block is at the equilibrium position, how much potential energy is in the springs? Where must the rest be?
 

1. How does the potential energy of a spring change as it is stretched or compressed?

The potential energy of a spring is directly proportional to the amount of stretch or compression it experiences. This means that as the spring is stretched, its potential energy increases and as it is compressed, its potential energy decreases. This relationship is described by Hooke's Law, which states that the force exerted by a spring is equal to the spring constant (k) multiplied by the distance the spring is stretched or compressed (x).

2. What factors affect the amount of potential energy stored in a spring?

The amount of potential energy stored in a spring is affected by two factors: the spring constant (k) and the distance the spring is stretched or compressed (x). The spring constant is a measure of the stiffness of the spring, and a higher spring constant means the spring will store more potential energy for a given amount of stretch or compression. The distance the spring is stretched or compressed also plays a role, as a longer stretch or compression will result in a greater potential energy.

3. How does the mass of an object attached to a spring affect its potential energy?

The mass of an object attached to a spring does not directly affect the potential energy of the spring. However, it does affect the amount of force required to stretch or compress the spring, which in turn affects the potential energy. A heavier object will require more force to stretch or compress the spring, resulting in a greater potential energy.

4. Can the potential energy of a spring be converted into other forms of energy?

Yes, the potential energy stored in a spring can be converted into other forms of energy, such as kinetic energy, when the spring is released. This is because the potential energy is a form of stored energy, and when the spring is released, this stored energy is transformed into the energy of motion as the spring returns to its original shape.

5. How does the potential energy of a spring affect its natural frequency?

The potential energy of a spring does not directly affect its natural frequency. The natural frequency of a spring is determined by its mass and stiffness, which are related to the potential energy through the spring constant. However, a spring with a higher potential energy will have a greater amplitude of oscillation, meaning it will stretch and compress more, resulting in a higher frequency of oscillation.

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