# How come you have to use work energy to find v0 of spring?

• I
• annamal
In summary: If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity. How come you cannot do kx*(delta t) = m*v and what would be the delta t value? Because in the first case you have the energy at a point ##x_1## (it's better to use this to see that it's a fixed point, not a variable). In the second case, ##x## is variable and the applied force varies with time. Because in the first case you have the energy at a point ##x_1## (it's better to use this to see that it's a fixed point, not a variable
annamal
TL;DR Summary
If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?
If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?

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annamal said:
TL;DR Summary: If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v and what would be the delta t value?

If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v and what would be the delta t value?
Because in the first case you have the energy at a point ##x_1## (it's better to use this to see that it's a fixed point, not a variable).

In the second case, ##x## is variable and the applied force varies with time.

PeroK said:
Because in the first case you have the energy at a point ##x_1## (it's better to use this to see that it's a fixed point, not a variable).

In the second case, ##x## is variable and the applied force varies with time.
I am just looking for the initial velocity though

annamal said:
I am just looking for the initial velocity though
The initial velocity is zero, I assume.

PeroK said:
The initial velocity is zero, I assume.
I mean what is the initial velocity once you let go of the spring pulled a certain distance x. Why can't you do that with kx*(delta t) and what is delta t

annamal said:
If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity
This is not correct, which may be the source of your confusion. A correct statement would be that ##\frac 12kx^2+\frac 12mv^2=\mathrm{const}##, which you can also get by considering forces if you want.

If you solve for the ##\mathrm{const}## you will find that it is ##\frac 12kx_{\mathrm{max}}^2##, where ##x_{\mathrm{max}}## is the maximum value of ##x##. If you then consider the case where ##x=0## you will find that this is where velocity is a maximum, which leads to ##\frac 12mv_{\mathrm{max}}^2=\frac 12kx_{\mathrm{max}}^2##, which may be what you are thinking of, but only determines the speed at equilibrium, not the initial speed.

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PeroK
annamal said:
I mean what is the initial velocity once you let go of the spring pulled a certain distance x.
The initial velocity is zero. The mass then accelerates over time, under a changing force.
annamal said:
Why can't you do that with kx*(delta t) and what is delta t
For small ##\Delta t##, we have ##F \approx kx_{max}## and ##mv \approx F\Delta t##. That only works for a short time, before you have to take the changing force into account.

Ibix
annamal said:
I mean what is the initial velocity once you let go of the spring pulled a certain distance x.
The initial velocity is whatever it's set to by the experimenter (perhaps zero). Your calculation with kx will tell you the Initial Acceleration of the mass and not the velocity

annamal said:
TL;DR Summary: If you use work energy, you can get 0.5*k*x^2 = 0.5*m*v^2 to get the velocity if you pulled the spring a distance x. How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?

How come you cannot do kx*(delta t) = m*v to get the initial velocity and what would be the delta t value?
The fact that you need to ask what delta t would be is the answer to 'why'. You want to use momentum instead of Energy so there must be a method to do it that way but using Potential Energy transfer to Kinetic Energy avoids all that hassle. If you do all that correctly, you would get the same (right) answer but life's too short.

The good thing about the Energy way is that Energy is conserved throughout the process.

vanhees71
Or to put it differently: The work-energy theorem holds for any force, i.e., from the equation of motion
$$m \dot{\vec{v}}=\vec{F},$$
where ##\vec{F}## can be an arbitrary function of ##\vec{x}##, ##t##, ##\dot{\vec{x}}##(, and even higher time derivatives of ##\vec{x}##, but this usually leads to other trouble). Now multiply this with ##\vec{v}## (using the scalar product). Then you get
$$m \vec{v} \cdot \dot{\vec{v}}=\vec{v} \cdot \vec{F}.$$
Now the left-hand side is a total time derivative, i.e.,
$$m \vec{v} \cdot \dot{\vec{v}}=\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 \right)=\dot{E}_{\text{kin}}=\vec{v} \cdot \vec{F}.$$
Now integrate this wrt. time over a time interval ##(t_1,t_2)## this gives the work-energy theorem,
$$E_{\text{kin}}(t_2)-E_{\text{kin}}(t_1)=\int_{t_1}^{t_2} \mathrm{d} t \vec{v} \cdot \vec{F}=W(t_1,t_2),$$
i.e., the change in kinetic energy is given by the work done in the said time interval.

This doesn't help much, if you don't know the solution of the equation of motion, and it's thus a pretty dull result. This drastically changes, if ##\vec{F}## is a potential field, i.e., if it's a function of ##\vec{x}## only, and it can be written as a gradient of a scalar field ##V##. Usually one defines
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Using this in the work-energy theorem leads to
$$W(t_1,t_2)=-\int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \cdot \vec{\nabla} V(\vec{x}) = -\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x})=-V(\vec{x}_2)+V(\vec{x}_1),$$
where ##\vec{x}_2=\vec{x}(t_2)## and ##\vec{x}_1=\vec{x}(t_1)##. So you have from the work energy theorem
$$E_1=E_{\text{kin} 1}+V(\vec{x}_1)=E_2=E_{\text{kin} 2} +V(\vec{x}_2),$$
i.e., the then definable total energy,
$$E=E_{\text{kin}}+V(\vec{x})=\text{const},$$
is conserved for any solution of the equation of motion, i.e., you don't need to know the specific equation of motion, to have some information from enery conservation.

For a force in 1D motion, ##F(x)=-k x## you have ##V(x)=k x^2/2## and thus
$$\frac{m}{2} v^2 + \frac{k}{2} x^2=\text{const}.$$
I'm not sure about the initial conditions in the OP. I guess, at ##t=0## you expanded the spring by ##x_0>0## from the equilibrium position ##x=0## and then realease it, i.e., your initial conditions are ##x(0)=x_0## and ##v(0)=0##. Then the energy-conservation law tells you that
$$E=\frac{m}{2} v^2 + \frac{k}{2} x^2=E_0=\frac{k}{2} x_0^2.$$
This tells you, without solving for the equations of motion that you always must have ##|x|<x_0## and that the maximal velocity is for ##x=0##, and that this maximal speed is given by
$$\frac{m}{2} v_{\text{max}}^2=E_0=\frac{k}{2} x_0^2 \; \Rightarrow \; |v_{\text{max}}|=\sqrt{\frac{k}{m}} x_0.$$
For ##|x|=x_0## you always have ##v=0##. This tells you that the mass oscillates between the points ##x=x_0## and ##x=-x_0##, and at this turning points ##\dot{x}=v=0##. So you get a pretty good qualitative picture about the motion, without solving for the equation of motion, just from the energy-conservation law, which holds for the forces, which have a potential in the above described sense.

sophiecentaur

## Why do we need to use work-energy principles to find the initial velocity of a spring?

Using work-energy principles allows us to account for the energy transformations that occur in the system. When dealing with springs, energy is stored as potential energy when the spring is compressed or stretched and converted to kinetic energy as the spring returns to its equilibrium position. This approach provides a straightforward way to relate the spring's displacement to its velocity.

## Can't we just use Newton's laws to find the initial velocity of the spring?

While Newton's laws can certainly be used to analyze the forces and accelerations involved, the work-energy principle offers a more direct method for solving problems involving energy transformations. It simplifies calculations by focusing on the conservation of energy rather than the detailed forces and motions at every point.

## How is the work done by the spring related to the initial velocity?

The work done by the spring is equal to the change in its potential energy, which is then converted into kinetic energy. By equating the work done (or energy stored) in the spring to the kinetic energy of the mass, we can solve for the initial velocity. Mathematically, this is represented as $$\frac{1}{2} k x^2 = \frac{1}{2} m v_0^2$$, where $$k$$ is the spring constant, $$x$$ is the displacement, $$m$$ is the mass, and $$v_0$$ is the initial velocity.

## What is the role of the spring constant in determining the initial velocity?

The spring constant $$k$$ measures the stiffness of the spring and determines how much force is needed to compress or stretch it by a certain amount. A higher spring constant means the spring stores more potential energy for a given displacement, which in turn affects the initial velocity when that energy is converted to kinetic energy. The initial velocity is directly proportional to the square root of the spring constant.

## Does the mass attached to the spring affect the initial velocity?

Yes, the mass attached to the spring affects the initial velocity. According to the work-energy principle, the kinetic energy of the mass is given by $$\frac{1}{2} m v_0^2$$. For a given amount of potential energy stored in the spring, a larger mass will result in a smaller initial velocity, as the energy is distributed over a greater mass. Conversely, a smaller mass will result in a higher initial velocity.

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