# Work-Energy Theorem for Moving Block and Spring System?

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• AyushNaman
In summary, the chapter discusses problems involving two block systems connected by a spring, with the goal of calculating the maximum elongation or compression in the spring. The most commonly used approach is the "Conservation of Mechanical Energy" method, but there is also a question about using the work-energy theorem. However, there is confusion about the expression for the work done by the spring on individual blocks in these questions, as the spring and blocks are both moving and their displacements are independent. The issue has left the writer confused and seeking clarification. The concept of switching to the center of mass frame is also discussed as an important concept in physics. However, for the 1D problem, the initial positions and length of the spring in equilibrium must be
AyushNaman
In the chapter of center of mass and linear momentum, there are multiple problems involving two block systems connected by a string, with both blocks given certain initial velocities. The goal is to calculate the maximum elongation or compression in the spring.

For example, consider this problem:
Two blocks of mass 3kg and 6kg are connected by a spring (stiffness=200), with initial velocities of 1m/s towards the left and 2m/s towards the right, respectively. To find the maximum elongation, the most commonly used approach is the "Conservation of Mechanical Energy" method. We calculate the velocity of the center of mass and equate the initial total kinetic energy to the later potential energy plus the later kinetic energy to find the 'x' term, representing the maximum elongation.

Now, the question is, how can one approach such questions using the work-energy theorem? I tried to apply it, but the main issue I encountered was figuring out the expression for the work done by the spring on individual blocks. This confusion arises from the following doubt:

During the derivation of the work done by a spring on a body, we took the spring to be attached to a certain support, such as a wall. Then the spring was connected to a block, displaced by an amount 'x', and then the basic integration followed :$$\int_{0}^{x}kx dx= \frac{kx²}{2}$$
However, the spring here chosen was at rest and the amount of force applied by the spring and the displacement of the block were both related by the 'x' term. What I mean is kx and the x in small displacement dx are the same, or in simple terms, the force on the block by the spring was a function of block's displacement and thus we could integrate.

HOWEVER, in these questions, the spring and blocks are both moving, and the elongation of the spring and the amount of a block's displacement are completely independent. A force kx(x-elongation in the spring) acts on both the blocks, but they don't displace by the same amount,i.e, the force by the spring the blocks now isn't a function on of their displacements, so we can't integrate and just write the regular {kx²/2} expression. Also, we can't simply integrate kx*ds without knowing the actual displacements of the blocks. Some might say it's 1/2k(x_1)^2 and 1/2k(x_2)^2 for individual blocks. But as I already described, the displacement and the elongation have no relation, so it's not that straightforward.

Also if the mechanical energy conservation approach is used, it is directly assumed that the potential energy in the spring is 1/2kx²(x being the total elongation in the spring). The proof of it is, in fact, based on the assumption that the spring is fixed, and the force acting on the spring is a function of its endpoint displacement. In this case, if one were to prove it from scratch, they cannot do so since both ends are moving with separate velocities, resulting in different displacements but under the same force, k * (elongation). So we would need to actually find out the work done by this conservative spring force.

This issue has left me confused, and despite asking my teacher and many others, no one could interpret my doubt. All the solutions provided to such questions are only using the conservation of mechanical energy. I would be utterly grateful if you could look into this and provide clarification.

Switching to the centre of mass frame is an important concept throughout physics. Personally, I can't see a good reason not to use that technique.

In fact, the sooner in your studies you fully embrace the principle of relativity the better!

In the centre of mass frame, you could to try to solve the problem using other methods, but conservation of energy is so Straightforward that why do anything else?

I don't know, how to answer the question, because the problem is stated incompletely. In addition to the initial velocities you also need the initial positions (or at least the initial relative position) and the length of the spring in equilibrium.

For the 1D problem (i.e. the two masses fixed on a rod and connected by the spring) the Hamiltonian in terms of the center-mass coordinates and momentum, ##(X,P)##, and relative-coordinates and momentum, ##(r,p)## is obvioulsy given by
$$H=\frac{\vec{P}^2}{2M} + \frac{p^2}{2 \mu} + \frac{k}{2} (r-l_0)^2 \quad \text{with} \quad M=m_1+m_2, \quad \mu=\frac{m_1 m_2}{M}.$$
Since ##H## is not explicitly time dependent, it's conserved, and since ##X## is cyclic, ##P=\text{const}##. To get the energy you can use ##P=P_0=m_1 v_1 + m_2 v_2## and ##p_0=\mu (v_1-v_2)##. In addition you need ##r_0##.

The "natural" assumption to make in this context is that r0= l0 but indeed it should be called out explicitly in the problem.
AyushNaman said:
In this case, if one were to prove it from scratch, they cannot do so since both ends are moving with separate velocities, resulting in different displacements but under the same force, k * (elongation). So we would need to actually find out the work done by this conservative spring force.
But you can see easilly that simply writing each position in terms of a center of mass position plus relative position will allow this to generalize. You should work this through if you don't see it.

vanhees71
AyushNaman said:
Also if the mechanical energy conservation approach is used, it is directly assumed that the potential energy in the spring is 1/2kx²(x being the total elongation in the spring). The proof of it is, in fact, based on the assumption that the spring is fixed, and the force acting on the spring is a function of its endpoint displacement. In this case, if one were to prove it from scratch, they cannot do so since both ends are moving with separate velocities, resulting in different displacements but under the same force, k * (elongation). So we would need to actually find out the work done by this conservative spring force.

For a massless spring, the net force acting on the spring must be zero at any instant. So, the force that mass ##m_1## exerts on the spring must be equal and opposite to the force that ##m_2## exerts on the spring. Newton’s third law implies that the force that ##m_1## exerts on the spring is equal and opposite to the force that the spring exerts on ##m_1##. Similarly for ##m_2##. Hence, the force ##F_{x1}## which the spring exerts on ##m_1## is always equal and opposite to the force ##F_{x2}## that the spring exerts on ##m_2##. If ##s## represents the amount of stretch of the spring from its unstretched configuration, then we have ##F_{x1} = ks## and ##F_{x2} = -ks##.

When ##m_1## moves ##dx_1## while ##m_2## moves ##dx_2##, the net work done on the two masses is $$dW = F_{x1}dx_1 + F_{x2}dx_2 = ks(dx_1-dx_2).$$ If ##l_0## is the unstretched length of the spring, then ##s = x_2 - x_1 - l_0##. So, ##ds = dx_2 - dx_1##. Thus, $$dW = -ksds.$$ Assuming that the spring is initially unstretched (##s = 0##) and that ##s_{max}## represents the maximum stretch of the spring, then the work done by the spring from the initial configuration to the instant the spring has maximum stretch is $$W = -\int_0^{s_{max}}ksds = -\frac 1 2 k s_{max}^2.$$ So, even though the two ends of the spring have different velocities and the work done on each mass is different, the net work done by the spring on the two masses is still given by ##-\frac 1 2 k s_{max}^2## in the lab frame of reference.

The work-energy theorem gives $$-\frac 1 2 ks_{max}^2 = \Delta K = \left(\frac 1 2 m_1 v_f^2 + \frac 1 2 m_2 v_f^2\right) - \left(\frac 1 2 m_1 v_{10}^2 + \frac 1 2 m_2 v_{20}^2\right)$$ Here, ##v_{10}## and ##v_{20}## are the initial velocities of ##m_1## and ##m_2##, respectively. [EDIT: Note that at the instant of maximum stretch, the two masses must have the same velocity ##v_f##.]

Since ##F_{1x} = -F_{2x}##, the rate of change of momentum of ##m_1## is always equal and opposite to the rate of change of momentum of ##m_2##. So, overall, the total momentum is conserved: $$m_1v_f+ m_2 v_f = m_1 v_{10} + m_2 v_{20}.$$ You can solve this for ##v_f## and substitute the result into the work-energy equation. Then you can solve for the maximum stretch of the spring, ##s_{max}##.

Last edited:
vanhees71

## What is the Work-Energy Theorem and how does it apply to a block and spring system?

The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. For a block and spring system, this means that the work done by the spring force as the block moves will result in a change in the block's kinetic energy. If the spring is compressed or stretched, the potential energy stored in the spring is converted into kinetic energy of the block and vice versa.

## How do you calculate the work done by a spring on a block?

The work done by a spring on a block is calculated using the formula $$W = \frac{1}{2} k x^2$$, where $$k$$ is the spring constant and $$x$$ is the displacement of the spring from its equilibrium position. This formula gives the elastic potential energy stored in the spring, which is equal to the work done by the spring when it returns to its equilibrium position.

## What is the role of potential energy in the Work-Energy Theorem for a block and spring system?

In a block and spring system, potential energy plays a crucial role in the Work-Energy Theorem. The potential energy stored in the spring when it is compressed or stretched is converted into kinetic energy of the block as it moves. The total mechanical energy (sum of kinetic and potential energy) of the system remains constant if there are no non-conservative forces (like friction) acting on the system.

## Can the Work-Energy Theorem be used for non-linear springs?

The Work-Energy Theorem can be applied to non-linear springs, but the calculations become more complex. For non-linear springs, the force is not directly proportional to the displacement, and the work done by the spring must be calculated using the integral of the force over the displacement. The basic principle of the theorem still holds: the work done by the spring results in a change in the kinetic energy of the block.

## How do external forces affect the application of the Work-Energy Theorem in a block and spring system?

External forces, such as friction or applied forces, affect the application of the Work-Energy Theorem by adding or removing energy from the system. If external forces are present, the work done by these forces must be accounted for in addition to the work done by the spring. The total work done by all forces will equal the change in the kinetic energy of the block. In the presence of non-conservative forces, the mechanical energy of the system is not conserved, and the Work-Energy Theorem must include the work done by these external forces.

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