D&D 5E Mage Hand and Trap Avoidance

So why 11.6 - infinity? Force is mass x acceleration. So the amount of Force It depends on how much acceleration it can impart. And how fast that thing was moving when it started, as it will need to accelerate/decelerate it appropriately.

Pounds are actually a unit of force, not mass. An object that, at rest, exerts 10lbs of force on a scale masses 0.31 slugs (imperial unit of mass) or 4.5kg (metric unit of mass)

For simplicity we'll assume continuous acceleration of an object that started at rest. The spherical cow of basic physics. May Newton guide my hand.

from d = 0.5(a)t^2 we get 30ft=0.5a(6s^2) => 1.67ft/s^2
::: 0.31 slug x (32.2ft/s^2 / + 1.67ft/s^2 ) = 11.6lbs

But wait! This is a floor! The actual force can be more! What if you catch a 0.31slug/4.5kg object that is falling? What if it got dropped last round and this round you are going to catch it with Mage Hand and then lift it 30ft?

the initial velocity will be v = at where a is gravity so 32.2ft/s^2 x 6s => -193ft/s
d=0.5at^2 ‐vt
30ft =a18s^2 -1158ft
66ft/s^2 = a
:::0.31slug x a=21lbs

Repeat up to v= terminal velocity.

but...this assumes a bit of "bounce" as the hand goes down on impact, then rises up. what if the hand is already at 30ft and can't move away? There's nothing saying anything can "push" a mage hand. More massive things just pass through. Less massive things...??

In that case you can convert the kinetic falling energy to force, that is e=F d, where f is the force and d is the distance the force is applied.

Except....if the hand isn't directed to move.....is the distance ~0?
Because if so, for e>0, F would approach infinity.
so...between 11.6lbs and infinity.

What you're looking for here is the impact force, which is calculated by the following formula:
F = Δp / Δt
where F is the impact force, Δp is the change in momentum, and Δt is the change in time.

So first, we calculate the change in momentum:
p = mv
where p is the momentum, m is the mass, and v is the velocity.

Velocity is easy to calculate: a mage hand can move up to 30 feet in 6 seconds, so that 5 feet per second (or 0.0015 m/s). The mass of a Mage Hand is not given, but from the description, I would assume the mass to be zero ("spectral") or lighter than air ("floating"). Let's be generous, and go with the mass of air. According to Google, a hand has a volume of about 410cc (or 0.00041 m³), and air has a density of 1.225kg/m³ at STP, so let's assume the mass of a Mage Hand is 5.02x10^-4 kg.

p = mv
p = (5.02x10^-4 kg)(0.0015 m/sec)
p = 7.65x10^-7 kg-m/s

Now to calculate the change in things (momentum and time), we need to define the initial and the final conditions. Initially, the mage hand is going to be at rest, 30 feet away from the pressure plate. Final condition will be the mage hand traveling at full speed and striking the plate.

Initial Condition: p = 0 because v = 0.
Final Condition: p = 7.65x10^-7 kg-m/s

Δp = p_final - p_initial
Δp = (7.65x10^-7 kg-m/s) - 0
Δp = 7.65x10^-7 kg-m/s

Now we need the impact time (or as it is sometimes called, the impact duration), which is the amount of time that the mage hand is in contact with that pressure plate. Assuming a perfectly inelastic collision (the mage hand doesn't stick to the plate), this would only be a fraction of a second. We calculate this with the following formula:
t₀ = sv_avg = 2sv_init,
where t₀ is the impact time, s is the distance over which the force is acting (the thickness of the pressure plate), and v_init is the velocity the mage hand is moving before it impacts the pressure plate.

So let's be generous, and say the pressure plate is only one millimeter thick (0.001 meters). Smaller impact time means larger force, so we are being conservative. If we use something thicker (like the lid of a chest, or a door), the impact time would be larger and the force would be less concentrated (and therefore lower). Anyway...

t₀ = 2(s)/v_init
t₀ = 2(0.001 m)/(1.524 m/sec)
t₀ = 0.0013 sec

Find the change in time from initial to final condition:

Δt = t₀ - t₁
Δt = 0.0013sec - 0 sec
Δt = 0.0013sec

And now we have Δp and Δt, so we can finally calculate F (the impact force).

F = Δp / Δt
F = (7.65x10^-7 kg-m/s) / (0.0013 sec)
F = 0.0000005022 kg-m/s²
F = 0.000588 Newtons
F = 0.000132 pounds(force)

That's about the amount of force a single grain of table salt would exert on a table.

So, not a lot.

First, convert weight to mass: 10 lbs(w) = 4.5359kg.
Then, solve for the new momentum: p = mv = (4.5359 kg)(0.0015m/sec) = 0.0068 kg-m/s
Finally, use the updated momentum to solve for force: F = Δp / Δt = (0.0068 kg-m/s)/0.0013 m/s) = 5.234 N = 1.17 lbs(force)

A little more than a pound of force. Way more then before (when the mage hand had the same mass as the air around it), but still not nearly as much as you would think. The velocity just isn't high enough.