Confused about some basic mechanics principles (force , PE)

[dyn]

Hi - i am confused about the following basic mechanics principles

Consider a large cubic mass ( mass M ) on the ground with a small cubic mass ( mass m ) placed on top of it. The surface between the 2 masses is frictionless

1 - if a push (apply a force ) to the large cube does the upper (small) cube remain stationary until the large cube is no longer below it and then fall ?

2 - if the large mass has an acceleration a and i apply Newton's 2nd law do i use F = Ma or F = ( M+m )a ?

If i now have the large cubic mass ( sides L and mass M) on the ground and a point mass ( mass m ) held at a height h above the ground. The potential energy (PE) of the point mass is mgh. Is the PE of the cubic mass Mg(L/2) because it is treated as point mass with the total mass at it's centre of mass ? If this is correct then it seems strange that a cube on the ground has potential energy because it cannot drop and so has no energy due to its position.

Thanks

 

[A.T.]

If this is correct then it seems strange that a cube on the ground has potential energy because it cannot drop and so has no energy due to its position.

It could melt, and then the potential energy is released.

But if you are sure that the some potential energy will stay constant in your scenario, then you don't have to consider it. Only the changes in potential energy matter.

 

[Lnewqban]

1) The top cube receives no horizontal force (there is no friction); therefore, nothing changes for it, regarding original momentum or velocity.
2) You only consider M, for same reason explained for 1) above.

The weight force is acting on the big block, which means it has the potential to do some vertical work on the block, if given the chance.
Until reaching the center of the planet, the block has some amount of gravitational energy to be used.
Ground surface is preventing vertical movement of the block; therefore, weight can't do any work

 

[kuruman]

If this is correct then it seems strange that a cube on the ground has potential energy because it cannot drop and so has no energy due to its position.

This argument is faulty. If there a hole through the center of the Earth to the other side and you dropped the mass it will fall in gaining kinetic energy, pass through the center of the Earth and eventually reach the other side with zero velocity at which point it will fall back in and so on. In other words it will oscillate back and forth inside the hole like a spring-mass system. If you hold a sretched spring with your hand and not allow it to move, this does mean that there is no potential energy stored in the spring. The same idea applies to mass resting on the surface of the Earth: the ground prevents it from moving.

The expression $U=mgh$ is an approximation and you need to understand where it's coming from. The potential energy of a system consisting of the Earth of mass $M_e$ and radius (assuming a perfect sphere) $R_e$ and an object of mass $m$ is given by $U=-\dfrac{GM_em}{r}$ where $r$ is the distance of the mass from the center of Earth. If the object is raised from distance $R_e$ to height $h$ from the surface of the Earth, the change in potential energy will be $$\Delta U=-\frac{GM_em}{R_e+h}-\left(-\frac{GM_em}{R_e}\right)=\frac{GM_e mh}{R_e(R_e+h)}.$$This expression is exact. The radius of the Earth is about 6.4 million meters. If $h$ is as large as hundreds of meters, it it still small relative to the radius of the Earth. Therefore, it is a good approximation to ignore $h$ in the denominator and write $\Delta U\approx\dfrac{GM_em h}{R_e^2}$. With the definition for the acceleration of gravity near the Earth's surface $g=\frac{GM_e}{R_e^2}$, you get the familiar $\Delta U=mgh$.

Note the following about $mgh$:
1. It is an approximation valid only near the Earth's surface, say at distances no higher than there is air to breathe.
2. It is a change in potential energy from an initial weight near the Earth's surface to a final height also near the Earth's surface. The initial height need not be at the surface of the Earth, but you have to define where it is and be consistent.

Finally, it should be obvious to you from all this that it is incorrect to think of a block as "having" potential energy. The potential energy is not a property of a single system but depends on the relative configuration of the components in a two-component system. Here, for example, the two components are the Earth and the mass $m$. If you change their relative position, you change their common potential energy. Without the Earth around, the mass can only have kinetic energy.

 

[dyn]

Thanks for your replies. I realize that PE =mgh is an an approximation and only valid near the Earth's surface. With basic mechanics question the zero of PE is taken to be at h=0 ; so it just seemed strange that a point mass could have zero PE but 3-D object wouldn't be able to have zero PE because it's centre of mass would always be above h=0

 

[kuruman]

Thanks for your replies. I realize that PE =mgh is an an approximation and only valid near the Earth's surface. With basic mechanics question the zero of PE is taken to be at h=0 ; so it just seemed strange that a point mass could have zero PE but 3-D object wouldn't be able to have zero PE because it's centre of mass would always be above h=0

The zero of potential energy is arbitrary as long as it is near the surface of the Earth. This means that you get to choose where to put it.
If you choose h = 0 to be at the center of mass of the 3-D object, the object will have zero PE.
What matters is the change in potential energy, not its value. It's as simple as that.

 

[Herman Trivilino]

I realize that PE =mgh is an an approximation and only valid near the Earth's surface. With basic mechanics question the zero of PE is taken to be at h=0 ; so it just seemed strange that a point mass could have zero PE but 3-D object wouldn't be able to have zero PE because it's centre of mass would always be above h=0

Only changes in potential energy have any physical meaning. (This is equivalent to being able to set the potential energy to zero at any location you choose.)