Ideal Spring, Spring Constant, Equilibrium Length

In summary, a 2.0 kg mass attached to one end of a spring with a spring constant of 500 N/m is compressed 10 cm on a frictionless surface. When released, the mass moves apart with a 4.0 kg mass attached to the other end. Using Hooke's law and work-energy theorem, it can be determined that the speed of the 2.0 kg mass when the spring reaches its equilibrium length is 1.0 m/s.
  • #1
Spartan Erik
31
0

Homework Statement



A 2.0 kg mass is attached to one end of an ideal spring with a spring constant of 500 N/m and a 4.0 kg mass is attached to the other end. The masses are placed on a horizontal frictionless surface and the spring is compressed 10 cm from its equilibrium length. The spring is then released with the masses at rest and the masses begin to move apart. When the spring has reached its equilibrium length, what is the speed of the 2.0 kg mass?

0.67, 1.0, 1.3, 2.1, none of these (all in m/s)

Homework Equations



Hooke's law: F = -kx
W(spring): 1/2kx(initial)^2 - 1/2kx(final)^2
W(applied): -W(spring) if stationary before/after displacement

The Attempt at a Solution



I don't know how I can utilize this data and relate it to velocity
 
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  • #2
Spartan Erik said:

Homework Statement



A 2.0 kg mass is attached to one end of an ideal spring with a spring constant of 500 N/m and a 4.0 kg mass is attached to the other end. The masses are placed on a horizontal frictionless surface and the spring is compressed 10 cm from its equilibrium length. The spring is then released with the masses at rest and the masses begin to move apart. When the spring has reached its equilibrium length, what is the speed of the 2.0 kg mass?

0.67, 1.0, 1.3, 2.1, none of these (all in m/s)

Homework Equations



Hooke's law: F = -kx
W(spring): 1/2kx(initial)^2 - 1/2kx(final)^2
W(applied): -W(spring) if stationary before/after displacement

The Attempt at a Solution



I don't know how I can utilize this data and relate it to velocity

OK. You kind of have the right idea.

The work that the spring produces has gone into what? (Does it rhyme with Kinetic Energy by any chance?)

Also you have two masses that are being acted on by the same force, even if in opposite directions.
F = m1a1 = m2a2

Since you know that m1 = 2 and m2 = 4, then you know that m2 = 2*m1

What does that mean then for a1 and a2? If that is true for a1 and a2 then it must also be true for v1 and v2 at the moment that PE is converted wholly to KE.

That leads to Work = total KE = ... you should be able to get the rest.
 
  • #3
of the 2.0 kg mass.

I would approach this problem by first understanding the concept of an ideal spring and its properties. An ideal spring is a theoretical construct that is assumed to have no mass, no damping, and no energy loss during oscillations. It is also assumed to obey Hooke's law, which states that the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium length.

In this problem, the spring constant (k) is given as 500 N/m, which means that for every 1 meter of displacement, the spring exerts a force of 500 Newtons. The equilibrium length of the spring is the length at which the spring is neither compressed nor stretched and has no force acting on it. In this case, the equilibrium length is not given, but it can be calculated using the formula F = -kx, where x is the displacement from equilibrium.

Now, let's consider the motion of the masses attached to the spring. When the spring is compressed 10 cm from its equilibrium length, it exerts a force on the masses in the opposite direction according to Hooke's law. This force causes the masses to accelerate in the direction of the force until the spring reaches its equilibrium length. At this point, the forces acting on the masses are balanced and they come to rest.

Using the conservation of energy principle, we can calculate the speed of the 2.0 kg mass when the spring reaches its equilibrium length. The potential energy stored in the spring when it is compressed can be calculated using the formula W(spring) = 1/2kx(initial)^2. Similarly, the potential energy stored in the spring when it reaches its equilibrium length can be calculated using the formula W(spring) = 1/2kx(final)^2. Since the total energy of the system (masses and spring) is conserved, the work done by the spring (W(spring)) must be equal to the work done by the applied force (W(applied)).

Substituting the values given in the problem, we get -1/2(500)(0.1)^2 = -1/2(500)x^2, where x is the displacement of the 2.0 kg mass from its equilibrium length. Solving for x, we get x = 0.05 m. This means that when the spring reaches its equilibrium length, the
 

1. What is an ideal spring?

An ideal spring is a theoretical concept in physics that represents a massless, frictionless, and perfectly elastic spring. It is used to simplify calculations and understand the behavior of real springs.

2. What is spring constant?

Spring constant, also known as force constant, is a measure of the stiffness of a spring. It is represented by the letter 'k' and is defined as the amount of force required to stretch or compress a spring by a unit distance.

3. How is equilibrium length of a spring determined?

The equilibrium length of a spring is the length at which it remains when no external forces are acting on it. It can be determined by hanging an object of known mass on the spring and measuring its length, then using the formula k = mg/x, where k is the spring constant, m is the mass, and x is the change in length.

4. What factors affect the ideal spring's behavior?

The behavior of an ideal spring is affected by its spring constant, mass attached to the spring, and the external forces applied to the spring. It also depends on the material and shape of the spring.

5. How does the ideal spring model differ from real springs?

Real springs have mass, experience friction, and have a limit to their elasticity, whereas ideal springs do not. Real springs also have a non-linear relationship between force and displacement, while ideal springs have a linear relationship. Additionally, real springs may have varying spring constants depending on the material and shape, while ideal springs have a constant spring constant.

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