There is a lot more to this question than initially meets the eye-- it is quite an interesting question.
Normally, in the context of fixed-wing flight, the thrust vector acts roughly parallel to the flight path through the airmass. When the thrust vector is exactly parallel to the flight path through the airmass, and the flight path is linear rather than curving up or down or to either side, then the vector diagram of forces in climb looks like this (left-hand case-- climb angle of 45 degrees-- right-hand case-- climb angle of 90 degrees):
Powered climb at climb angles of 45 and 90 degrees:
We can see that Lift = Weight * cosine (climb angle). In the left-hand diagram, the climb angle is 45 degrees and Lift = .707 * Weight. In the right-hand diagram, the climb angle is 90 degrees and Lift is zero.
But, these diagrams assume that the Thrust vector acts parallel to the flight path through the airmass. Obviously, if this isn't true, thrust equation lift = weight * cosine (climb angle) is also no longer true. To take an extreme case, note that when the exhaust nozzles of a Harrier "jump jet" are pointed straight down, the wing is "unloaded"-- the plane can hover at zero airspeed with zero lift, supported entirely by thrust. Conversely, during a glider winch launch, the towline pulls steeply downward on the glider. This too can be viewed as a form of "vectored thrust"-- but now the load on the wing is increased, rather than decreased, so the wings must generate a lift force that is much greater than the aircraft's weight.
In the case presented in this question, Thrust does NOT act along the flight path of the "aircraft", if we consider the wing to be the "aircraft". When the wing is rising up the pole, the vertical motion causes a change in the direction of the wing's trajectory through the airmass and also a change in the direction of the "relative wind", but there is no corresponding change in the direction of the Thrust vector. Thus the thought experiment presented in the question is NOT representative of the typical situation in fixed-wing flight. The thrust vector is NOT fixed in direction relative to the chord line of the wing, and is NOT acting roughly parallel to the direction of the "relative wind" experienced by the wing, and the direction of the wing's flight path through the airmass.
Furthermore, the basic mechanism that governs the airspeed of a fixed-wing aircraft is absent. Normally, as an aircraft climbs, if Lift exceeds Weight, the flight path will curve upward, causing the Weight vector to have a greater component acting parallel to the direction of the aircraft's flight path through the airmass, causing a decrease in speed. But in this experiment, since the wing is "locked" into position on the cart in the fore-and-aft sense, if the wing's trajectory curves upward, it appears that the cart will provide however much thrust is needed to hold the horizontal component of the wing's velocity vector component exactly constant. Assuming, that is, that the wing's drag is trivial compared to drag from other sources such wheel drag and wheel bearing drag from the cart, so that variations in drag from the wing have essentially no effect on the airspeed and groundspeed of the cart.
So the forces acting on the wing in this thought experiment will be very different from the forces typically acting on a fixed-wing aircraft in actual flight. It should not come as surprise to discover that in the case of this thought experiment, lift actually must be greater than weight in order for the wing to climb up the pole.
We really could end this answer right here. But it's rather interesting to look a little more deeply at the forces acting on the wing in the thought experiment.
What are some of the notable features of the thought experiment?
As we've already noted, the wing is locked in place relative to the cart in the fore-and-aft direction. The wing cannot speed up or slow down in the fore-and-aft direction, relative to the cart. Furthermore, IF the drag from the wing is trival compared to the other sources of drag acting on the cart, so that the drag from the wing has essentially no effect on the cart's airspeed, this means that the wing cannot speed up or slow down in the fore-and-aft direction relative to the airmass (or relative to the ground). The cart will transmit to the wing however much thrust is needed to hold constant the fore-and-aft component of the wing's airspeed vector. This is very different from the typical situation in fixed-wing flight.
Also. as the thought experiment was originally worded, the wing is locked into a constant pitch attitude.
Here is a vector diagram illustrating the situation when the cart is moving at some constant airspeed that is NOT high enough to allow the wing to lift off the cart:
The forces illustrated include Lift (L), Drag (D), Weight (W), Thrust (T), and the upward force (C) exerted by the cart on the wing as the wing rests on the cart. Net force is zero. The L/D ratio is 10:1.
Now assume that we hold the wing down with a catch as we increase power and accelerate to some higher airspeed, and then allow everything to stabilize. Then we unlatch the catch. The diagram below shows the situation at the instant we unlatch the catch--
The wing has not yet begun to rise upward, so there is no change in the direction of the wing's trajectory through the airmass, or the direction of the lift and drag vectors. The wing's angle-of-attack has not changed, so the lift and drag coefficients have not changed, so the L/D ratio is still 10/1. The dashed line represents the net force vector, which is simply the vector sum of all the other force vectors. Acceleration = force / mass, so we can also label the net force vector as "Acceleration * mass".
What happens as the wing starts to rise (accelerate) up the pole? The wing's upward velocity causes a change in the "relative wind" experienced by the wing. The wing's angle-of-attack immediately decreases or becomes negative, so the lift coefficient decreases, and the L/D ratio decreases. (The drag coefficient might decrease too, but not as much as the lift coefficient.) If the wing has a cambered, non-symmetrical airfoil, it will still produce lift at some small negative angle-of-attack, but not very much-- the lift coefficient will be low. When the wing reaches some given upward vertical velocity, Lift will have decreased to the point such that the vertical component of the net aerodynamic force acting on the wing will no longer be larger than Weight, the net force on the wing will drop to zero, and the wing will no longer be able to accelerate, but rather will move up the pole with a constant vertical velocity. The figure below illustrates this situation:
The direction of the wing's path through the airmass is parallel to (and opposite to) the direction of the Drag vector. The climb angle-- labelled "C" in the diagram-- is the acute angle formed between the Thrust and Drag vectors, and also between the Lift and Weight vectors. This is also the angle between the Drag vector and the horizon, and also the angle between the Lift vector and the vertical direction. The vectors can be all arranged head-to-tail in a closed figure, so the net force is zero. Lift is slightly larger than Weight, and Thrust is quite a bit greater than Drag. If the wing is mounted on the cart with zero incidence, then the wing's angle-of-attack must be slightly negative-- in fact it must be equal to negative "C" degrees. We've drawn the L/D ratio as 2/1, to represent the decrease in the wing's lift coefficient caused by the change in angle-of-attack. The wing is moving up the pole at a constant velocity.
Interestingly, this situation is virtually identical to the situation experienced by the rising wing as an aircraft rolls to a steeper bank angle, especially if the roll is driven by a spoiler deployed on the descending wing with no modification to the shape of the rising wing. The change in angle-of-attack caused by the wing's rising motion through the airmass limits the vertical speed that the rising wing can attain-- this is called "roll damping". The lift and drag vectors are "twisted aft" or "twisted backwards" from the direction they pointed before the rolling motion started-- you can also see this "twist" illustrated in this diagram https://www.av8n.com/how/img48/adverse-yaw-steady.png from this section https://www.av8n.com/how/htm/yaw.html#sec-adverse-yaw of the excellent "See How It Flies" website https://www.av8n.com/how/ .
The situation is also exactly like the situation we'd have if we were towing a glider under the following conditions: 1) We have a very long tow rope-- so long that the angle of the tow rope relative to the horizon is not influenced at all by the glider's climb rate relative to the tow plane. 2) The towplane is flying in such direction such that the glider's end of the rope is pulling exactly horizontally on the glider 3) The glider's drag is trivial compared to the towplane's thrust and drag, so the glider's aerodynamic situation has no influence on the towplane's airspeed. 4) The glider pilot is giving pitch control inputs in such a way as to force the glider's pitch attitude to stay constant relative to the horizon, regardless of climb rate.
Now, what if we modify the experiment by allowing the wing's pitch attitude to vary, while holding the wing's angle-of-attack constant relative to the wing's trajectory through the airmass-- perhaps by adding a stabilizing vane or tail to the back of the wing?
Now what happens when the wing starts to rise?
The figure below illustrates a situation where lift is exactly equal to weight. The drag vector is horizontal, so the wing cannot be rising or falling through the air-- it must be staying in a fixed position on the pole.
Note that we've chosen to illustrate a 5:1 L/D ratio for this version of the thought experiment.
Now what if we give the wing the tiniest upward push, so that it starts to rise? As soon as it starts to rise, its velocity through the airmass is augmented by its vertical motion. And now the wing is free to pivot in such a way that its angle-of-attack can stay constant, so we don't have the "damping" effect that we had in the earlier version of the experiment. The resulting increase in airspeed and lift is much like we see when a glider rises on a winch tow, except that in the case of the wing on the imaginary cart, the thrust vector stays horizontal, rather than pointing partly downwards. The wing's climb angle through the airmass will get steeper and steeper as its vertical speed increases. This causes the airspeed to increase, which causes the Lift vector to get larger.
The figure below illustrates the situation that we see when the wing's climb angle through the airmass reaches 60 degrees. Again, the dashed line represents the net force vector, which is simply the vector sum of all the other force vectors. Acceleration = force / mass, so we can also label the net force vector as "Acceleration * mass".
In this particular case, we've sized the lift and weight vectors to represent the situation where the horizontal component of the wing's speed through the airmass is exactly the same as it was in the previous diagram above, where lift was exactly equal to weight in the case where the wing's trajectory was horizontal. Simply by rising upward, the wing has experienced a doubling of airspeed and a four-fold increase in the magnitude of the lift vector. The sum of the vertical components of the lift and drag vectors is now 1.3 times the weight of the wing. Of course, we could modify the diagram to represent a case where the wing was experiencing a net upward force even before it started accelerating upward, simply by decreasing the size of the weight vector relative to the other vectors.
If the cart's velocity stays constant, and cart is able to transfer however much thrust is needed to the wing to keep it locked in place in the fore-and-aft direction relative to the cart, will the wing keep accelerating faster and faster up the pole?
It turns out it will not. Even if the wing has no weight at all, it will stop accelerating upwards once its climb angle equals the inverse tangent of the L/D ratio. For the 5/1 case illustrated here, that climb angle is 78.7 degrees. If the wing does have weight, the maximum achievable climb angle will be less. In the particular case illustrated above, where the Weight vector is exactly equal to the Lift vector that existed when the wing had zero upward velocity, the maximum achievable climb angle is somewhere between 70 and 75 degrees. Above this maximum achievable climb angle, the vertical components of Lift and Drag no longer add up to a value that is greater than Weight. So even when the wing is free to pivot to maintain a constant angle-of-attack, and the cart has infinite thrust available to allow it to maintain a constant airspeed in spite of changes in the wing's drag force, there is a limit to the climb angle that the wing can achieve.
Here's an interesting table
ca- cos- sin- aspd- L- D- vcL- vcD- net aero vc- net vert
0 1.00 0.00 1.00 1.00 .200 1.00 0.00 1.00 0.00
30 .866 .500 1.15 1.33 .267 1.15 .133 1.02 .021
45 .707 .707 1.41 2.00 .400 1.41 .283 1.13 0.13
60 .500 .866 2.00 4.00 .800 2.00 .693 1.31 0.31
70 .342 .940 2.92 8.55 1.71 2.92 1.61 1.32 0.32
75 .259 .966 3.86 14.9 2.99 3.86 2.88 .980 -0.02
80 .174 .985 5.76 33.2 6.63 5.76 6.53 -.773 -1.773
Constant horizontal component of airspeed in all cases
5/1 L/D ratio.
The last column (but only the last column) assumes that the value of Weight is selected in such a way that Weight is exactly equal to the value of the Lift vector in the particular case where the climb angle is zero, meaning that Weight and Lift are exactly in balance in this case. The value of Weight has no effect on any of the other columns.
ca= climb angle in degrees
cos= cosine of climb angle
sin= sine of climb angle
airspd= speed of wing through air in arbitrary units
L= Lift in arbitrary units
D= Drag in the same units as L
We've chosen an L/D ratio of 5/1
vcL= vertical component of lift (acts upwards) = L * cosine (climb angle)
vcD= vertical component of drag (acts downwards) = D * sine (climb angle)
net aero vc = vertical component of net aerodynamic force= (vcL-vcD) -- a positive sign means that the net aerodynamic force acts upwards, while a negative sign means that the net aerodynamic force acts downwards.
net vert = net vertical force = (net aero vc - weight), assuming that Weight is selected in such a way that Weight is exactly equal to the value of the Lift vector in the case where the climb angle is zero.
If the last column (net vert) is negative, this means that in the case where Weight is set to the particular value described above, the climb rate must decelerate (and the climb angle must decrease).
If the second-to-last column is negative, the climb rate must decelerate (and the climb angle must decrease) even if Weight is zero.
This version of the thought experiment-- where the pitch attitude of the wing is free to vary to maintain a constant angle-of-attack-- is somewhat like what happens at the start of glider winch tow, especially near the beginning of the tow when the towline is very long and the tow force stays almost horizontal for a while, even if the glider starts to climb rapidly.
Finally-- the original question contained the following line: " Please notice that - because thrust is horizontal - the chemical energy burned goes into kinetic energy of the cart and/or heat energy (due to overcoming drag). No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift." Thrust certainly does do work along the direction of the flight path through the airmass, which is never purely vertical. The situation appears to be analogous to a lightweight cube (say made of balsa wood) being blown up a slippery ramp by a wind that is blowing horizontally. Is the wind increasing the potential energy of the cube?
For more on the more "conventional" climbing situation-- a fixed-wing aircraft with thrust acting parallel to the direction of the flight path-- see these related answers to related questions:
:Does lift equal weight in a climb?"
"What produces Thrust along the line of flight in a glider?"
"'Gravitational' power vs. engine power"
"Descending on a given glide slope (e.g. ILS) at a given airspeed— is the size of the lift vector different in headwind versus tailwind?"
"Are we changing the angle of attack by changing the pitch of an aircraft?"