Solving the Block-on-Spring Problem Using Forces

[fog37]

Hello,

The classic problem of a block dropped on a vertical spring from a height $h$ above the scale: find the overall compression distance of a spring when a block is dropped on it and brought to rest. This problem is easy to solve using conservation of energy, potential gravitational energy and potential elastic energy. This is the energy approach.

What if we tried to solve the problem only using forces and body diagrams? I have been stuck thinking about it. How would we set the problem up? When the spring is fully compressed, the forces involved would be the weight of the block $W$ pointing down, the spring force $F_s = k \Delta x$ pointing up and the normal force $F_N$ pointing up.

We can first find the velocity of impact of the block with the spring: $$v_f = v_0 -2*9.8 (h) = -19.6 (h)$$ Then we can find the deceleration $a_{net}$ that the block experience going from $v_f$ to $0$.
Third, we can set up the force equation $$F_{net} = m a_{net} =F_N + k \Delta x - mg$$
We know $a_{net}$, $g$, $k$, $m$, but we don't know the normal force $F_N$ so we cannot find $\Delta x$...

Using the energy approach, the mass $m$ does not matter...

Thanks for any hint.

 

[BvU]

When the spring is fully compressed, the forces involved would be the weight of the block W pointing down, the spring force $F_s=k\Delta x$ pointing up and the normal force $F_N$ pointing up.

Seems to me you have one too many. Consider which force is working on which part of the system.

For an ideal spring that stays in place the forces working on each end are the same but opposite.

$\$

 

[erobz]

Using the energy approach, the mass m does not matter...

Also, care to elaborate on this?

And another thing you'll find is that even when considering the forces, the most efficient solution for how far the spring compresses will be "The Energy" argument in disguise. You will come to this by changing the independent variables in Newtons 2^nd Law formulation from time $t$ to compression $x$ through the application of the Chain Rule.

You can go to $x(t)$ directly, but it requires knowledge on the solution techniques to 2^nd order non-homogeneous linear ODE's with constant coefficients.

 

[hutchphd]

What is F_N (i.e what causes it)? Also the result will depend upon both m and k. Using Newtons Force law correctly will result in a (differential) equation whose answer will be more difficult but identical to Energy conservation.

 

[fog37]

What is F_N (i.e what causes it)? Also the result will depend upon both m and k. Using Newtons Force law correctly will result in a (differential) equation whose answer will be more difficult but identical to Energy conservation.

So $F_N$ represents the normal force, i.e. the contact force that the surface of the spring exerts on the block touching it. The spring also exert another force, the resistive force $F_s$ whose magnitude is equal to $k \Delta x$. Maybe the normal force $F_N$ is actually the same as $F_s$ so I can remove $F_N$ from my Newton's 2nd law equation...

The energy approach would give that, for a block of mass $m$ dropped from a height $h$ above the spring top surface, the gravitational potential energy $mg(h+\Delta x)$ becomes equal to the elastic potential energy $\frac {1} {2} k \Delta x^2$: $$mg(h+\Delta x)=\frac {1} {2} k \Delta x^2$$

My mistake, the mass $m$ does matter. The zero grav potential energy reference position is where the spring is fully compressed...

 

[fog37]

Also, care to elaborate on this?

And another thing you'll find is that even when considering the forces, the most efficient solution for how far the spring compresses will be "The Energy" argument in disguise. You will come to this by changing the independent variables in Newtons 2^nd Law formulation from time $t$ to compression $x$ through the application of the Chain Rule.

You can go to $x(t)$ directly, but it requires knowledge on the solution techniques to 2^nd order non-homogeneous linear ODE's with constant coefficients.

Hi erobz,

Thank you. So the only forces acting on the block are the weight $W$ and the resistive spring force $F_s$? I am conditioned to think that when a solid object box) touches another solid object (spring), a normal contact force $F_N$ must be included...But it sounds wrong in this scenario...

 

[erobz]

Hi erobz,

Thank you. So the only forces acting on the block are the weight $W$ and the resistive spring force $F_s$? I am conditioned to think that when a solid object box) touches another solid object (spring), a normal contact force $F_N$ must be included...But it sounds wrong in this scenario...

Hello!

Yeah, its a conditioning problem. The normal force is there, its just that it appears in both the FBD of the mass, and the FBD of the spring. As you eluded to, we are typically only looking at this thing from one side when we consider the contacting body ( the ground for instance ) to be rigid and kind of "unresponsive" to loading as far as changing its own geometry goes.

I think in reality everything is actually a spring to some extent, and there are no truly rigid things, but we approximate when it really doesn't make that much of a difference. I think... Maybe not having that explained causes some conceptual confusion.

So please feel free to continue on, ( I'm assuming you have knowledge of Calculus ) because we aren't quite there yet as far as finding your $x$ is concerned from Newtons Second Law.

 

[fog37]

Hello!

Yeah, its a conditioning problem. The normal force is there, its just that it appears in both the FBD of the mass, and the FBD of the spring. As you eluded to, we are typically only looking at this thing from one side when we consider the contacting body ( the ground for instance ) to be rigid and kind of "unresponsive" to loading as far as changing its own geometry goes.

I think in reality everything is actually a spring to some extent, and there are no truly rigid things, but we approximate when it really doesn't make that much of a difference. I think... Maybe not having that explained causes some conceptual confusion.

So please feel free to continue on, ( I'm assuming you have knowledge of Calculus ) because we aren't quite there yet as far as finding your $x$ is concerned from Newtons Second Law.

View attachment 301272

Well, the ODE to solve seems to be $$m \frac {d^2 x} {dt^2} = -mg + kx$$ with the additional conditions $\frac {dx} {dt} =0$ and maybe $x(t_0)=0$...

I need to brush up on my calculus to solve this ODE...I can do it numerically I guess...

 

[erobz]

Well, the ODE to solve seems to be $$m \frac {d^2 x} {dt}^2 = -mg + kx$$ with the additional conditions $\frac {dx} {dt} =0$ and maybe $x(t_0)=0$...

I need to brush up on my calculus to solve this ODE...I can do it numerically I guess...

Just for consistency using the convention in the diagram ( yours is fine for your convention, but opposite the diagrams convention ).

You are correct.

$$ m \frac{d^2x}{dt^2} = -kx + mg \tag{1}$$

with initial conditions: $x(0) = 0$ and $x'(0) = v_o$

Lets not worry about brushing up on the 2nd Order ODE solution just yet. Insted let's start with the equivalent statement:

$$ m \frac{dv}{dt} = -kx + mg \tag{1'} $$

By applying the Chain Rule to the LHS we:

$$ \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v $$

Substitute that into (1'):

$$ m\frac{dv}{dx} v = -kx + mg \tag{2} $$

The equation has been transformed into a first order separable ODE.

$$ \int_{v_o}^{0} mv \, dv = \int_{0}^{x_{max}} \left( -kx + mg \right) \, dx \tag{3} $$

After integrating (and moving some terms around), (3) should look very familiar! Also, it satisfies your initial question without "beginning" with CoE.

If you feel you've been cheated we can do the 2nd order ODE too!

 

[fog37]

Wow! Thanks erobz. This proves that using energy (kinetics) to solve problems is much more convenient than using forces (dynamics).

Since I have your attention here, in regards to potential energy (whatever type it may be), my definition is that it is the energy that a system formed by 2 or more components possesses in virtue of the fact the components are in a relative spatial position (configuration). A system having only a single component (ex: a point mass) does not have potential energy. However, a single extended object, since it is formed by multiple parts, can have potential energy...

Also, it is said that potential energy can only be defined in terms of a conservative force whose work, as line integral, is path independent. My understanding is that a non-conservative force would be path dependent which means we could go from a configuration $C_1$ to a different configuration $C_2$ in different ways. Given $PE_{C_1}$ and $PE_{C_2}$, the change in potential energy $\Delta PE= PE_{C_2} -PE_{C_1}$ between the two configurations would then not have a single unique value and that would be a problem...

That's why only the work done by conservative forces can set to be equal to the change in $PE$. That does not mean that a nonconservative force cannot contribute to changing the $PE$ of a system in a particular configuration, correct? For example, the force applied by my hand when it raises a book produces mechanical work which can be larger or equal to the change in gravitational potential energy. However, the work done by conservative forces is always exactly equal to $\Delta PE$...

Any flaw with my thinking?

 

[erobz]

I'm not a physicist (just an MET). So I'm going to respectfully bow out on "physics definitions" type inquiries. However, if none reply I'll try to work through it with you.

 

[erobz]

Wow! Thanks erobz. This proves that using energy (kinetics) to solve problems is much more convenient than using forces (dynamics).

Well, it depends on what you're after. It was convenient for this particular result, but if you are after $x(t)$ it all boils down to the same D.E. whether you start with Lagrangian or Hamiltonian formulations that use energy as the starting point. That being said, There was a reason why the other mechanics were developed, and sure there is another class of problems above my head where the distinction is important.