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Your question has some hidden assumptions. For example I could "lift" the object without applying any force at all – if the object already has an initial upward velocity. Of course you might say "this isn't what I meant by 'lifting'", but this shows the necessity of making questions and statements as clear as possible.

When we speak about "work" we must specify (1) which force is doing the work, (2) in which reference frame the work is measured. In fact two observers in two different frames may assign different works to the same force, because they will measure different velocities. For example the object may be at rest with respect to one observer, so no force is doing any work for that observer. (Note that all observers, inertial and non-inertial, agree on the values of the forces: forces are frame-invariant quantities.)

The work done on a point-like object during a time interval $[t_0,t_1]$ by the force $\pmb{F}(t)$ in a reference frame in which the object has velocity $\pmb{v}(t)$ is $$\int_{t_0}^{t_1} \pmb{F}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$ As the notation indicates, force and velocity may have an arbitrary time dependence. The product $\pmb{F}(t)\cdot \pmb{v}(t)$ is called the power expended by the force $\pmb{F}$ at time $t$ in that reference frame.

Note that the velocity is determined not just by the force $\pmb{F}$, but by all forces acting on the object during the time interval, and by the initial kinematic conditions, such as initial velocity.

According to Newton's laws, the sum of the forces $\pmb{F}^{(1)}, \pmb{F}^{(2)}, \dotsc$ acting on the object must equal, in an inertial frame (such as approximately the one fixed with the floor), the rate of change of momentum in that frame: $$\frac{\mathrm{d} (m \pmb{v})}{\mathrm{d}t} = \sum_{k} \pmb{F}^{(k)}(t)\ .$$

Assuming that the object isn't losing or acquiring mass, if we scalar-multiply this equation by $\pmb{v}$ and time-integrate between $t_0$ and $t_1$ we find $$\tfrac{1}{2}m\, v(t_1)^2 - \tfrac{1}{2}m\, v(t_0)^2 = \sum_k \int_{t_0}^{t_1} \pmb{F}^{(k)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$

For the gravitational force $\pmb{F}^{(\text{grav})} := -m\pmb{g}$ (with $\pmb{g}$ directed upwards) we also find, by simple integration, that $\int_{t_0}^{t_1} \pmb{F}^{(\text{grav})}(t)\cdot \pmb{v}(t)\ \mathrm{d}t = -mgh$, where $h$ is the vertical component of the total displacement of the object, considered positive if upward. We can note again that this displacement is observer-dependent: if we fix a camera on the lifted object, then with respect to the camera the object isn't moving, so $h=0$ in the reference frame of the camera, and the gravitational force hasn't done any work in that frame.

Let's now assume that the object is at rest (in the inertial frame of the floor) at times $t_0$ and $t_1$; that's probably implicit in the idea of "lifting". Then the left side of the equation above is zero. So the right side must also be zero. If the forces acting on the object are gravity and only another one, then from the results so far we have that the work done by the other force during this interval in this frame must be $+mgh$.

If we have gravity and two other forces acting on the object, we have that the total work done by the two extra forces together must be $+mgh$. But each force can have done an amount of work different from this. And so on for more forces.

There are many good books about such matters. For example Synge & Griffith's Principles of Mechanics, or Love's Theoretical Mechanics, or Truesdell's A First Course in Rational Continuum Mechanics, which treats all such matters in great depth and with logical care.

Your question has some hidden assumptions. For example I could "lift" the object without applying any force at all – if the object already has an initial upward velocity. Of course you might say "this isn't what I meant by 'lifting'", but this shows the necessity of making questions and statements as clear as possible.

When we speak about "work" we must specify (1) which force is doing the work, (2) in which reference frame the work is measured. In fact two observers in two different frames may assign different works to the same force, because they will measure different velocities. For example the object may be at rest with respect to one observer, so no force is doing any work for that observer. (Note that all observers, inertial and non-inertial, agree on the values of the forces: forces are frame-invariant quantities.)

The work done on a point-like object during a time interval $[t_0,t_1]$ by the force $\pmb{F}(t)$ in a reference frame in which the object has velocity $\pmb{v}(t)$ is $$\int_{t_0}^{t_1} \pmb{F}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$ As the notation indicates, force and velocity may have an arbitrary time dependence. The product $\pmb{F}(t)\cdot \pmb{v}(t)$ is called the power expended by the force $\pmb{F}$ at time $t$ in that reference frame.

Note that the velocity is determined not just by the force $\pmb{F}$, but by all forces acting on the object during the time interval, and by the initial kinematic conditions, such as initial velocity.

According to Newton's laws, the sum of the forces $\pmb{F}^{(1)}, \pmb{F}^{(2)}, \dotsc$ acting on the object must equal, in an inertial frame (such as approximately the one fixed with the floor), the rate of change of momentum in that frame: $$\frac{\mathrm{d} (m \pmb{v})}{\mathrm{d}t} = \sum_{k} \pmb{F}^{(k)}(t) \equiv \pmb{F}^{(1)}(t) + \pmb{F}^{(2)}(t) + \pmb{F}^{(3)}(t) + \dotsb\ .$$

Assuming that the object isn't losing or acquiring mass, if we scalar-multiply this equation by $\pmb{v}$ and time-integrate between $t_0$ and $t_1$ we find $$\begin{split}\tfrac{1}{2}m\, v(t_1)^2 - \tfrac{1}{2}m\, v(t_0)^2 &= \sum_k \int_{t_0}^{t_1} \pmb{F}^{(k)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t \\[1em]&\equiv \int_{t_0}^{t_1} \pmb{F}^{(1)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t + \int_{t_0}^{t_1} \pmb{F}^{(2)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t + \int_{t_0}^{t_1} \pmb{F}^{(3)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t + \dotsb \ .\end{split}$$

For the gravitational force $\pmb{F}^{(\text{grav})} := -m\pmb{g}$ (with $\pmb{g}$ directed upwards) we also find, by simple integration, that $\int_{t_0}^{t_1} \pmb{F}^{(\text{grav})}(t)\cdot \pmb{v}(t)\ \mathrm{d}t = -mgh$, where $h$ is the vertical component of the total displacement of the object, considered positive if upward. We can note again that this displacement is observer-dependent: if we fix a camera on the lifted object, then with respect to the camera the object isn't moving, so $h=0$ in the reference frame of the camera, and the gravitational force hasn't done any work in that frame.

Let's now assume that the object is at rest (in the inertial frame of the floor) at times $t_0$ and $t_1$; that's probably implicit in the idea of "lifting". Then the left side of the equation above is zero. So the right side must also be zero. If the forces acting on the object are gravity and only another one, then from the results so far we have that the work done by the other force during this interval in this frame must be $+mgh$.

If we have gravity and two other forces acting on the object, we have that the total work done by the two extra forces together must be $+mgh$. But each force can have done an amount of work different from this. And so on for more forces.

There are many good books about such matters. For example Synge & Griffith's Principles of Mechanics, or Love's Theoretical Mechanics, or Truesdell's A First Course in Rational Continuum Mechanics, which treats all such matters in great depth and with logical care.

Your question has some hidden assumptions. For example I could "lift" the object without applying any force at all – if the object already has an initial upward velocity. Of course you might say "this isn't what I meant by 'lifting'", but this shows the necessity of making questions and statements as clear as possible.

When we speak about "work" we must specify (1) which force is doing the work, (2) in which reference frame the work is measured. In fact two observers in two different frames may assign different works to the same force, because they will measure different velocities. For example the object may be at rest with respect to one observer, so no force is doing any work for that observer. (Note that all observers, inertial and non-inertial, agree on the values of the forces: forces are frame-invariant quantities.)

The work done on a point-like object during a time interval $[t_0,t_1]$ by the force $\pmb{F}(t)$ in a reference frame in which the object has velocity $\pmb{v}(t)$ is $$\int_{t_0}^{t_1} \pmb{F}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$ As the notation indicates, force and velocity may have an arbitrary time dependence. The product $\pmb{F}(t)\cdot \pmb{v}(t)$ is called the power expended by the force $\pmb{F}$ at time $t$ in that reference frame.

Note that the velocity is determined not just by the force $\pmb{F}$, but by all forces acting on the object during the time interval, and by the initial kinematic conditions, such as initial velocity.

According to Newton's laws, the sum of the forces $\pmb{F}^{(1)}, \pmb{F}^{(2)}, \dotsc$ acting on the object must equal, in an inertial frame, the rate of change of momentum in that frame: $$\frac{\mathrm{d} (m \pmb{v})}{\mathrm{d}t} = \sum_{k} \pmb{F}^{(k)}(t)\ .$$

Assuming that the object isn't losing or acquiring mass, if we scalar-multiply this equation by $\pmb{v}$ and time-integrate between $t_0$ and $t_1$ we find $$\tfrac{1}{2}m\, v(t_1)^2 - \tfrac{1}{2}m\, v(t_0)^2 = \sum_k \int_{t_0}^{t_1} \pmb{F}^{(k)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$

For the gravitational force $\pmb{F}^{(\text{grav})} := -m\pmb{g}$ (with $\pmb{g}$ directed upwards) we also find, by simple integration, that $\int_{t_0}^{t_1} \pmb{F}^{(\text{grav})}(t)\cdot \pmb{v}(t)\ \mathrm{d}t = -mgh$, where $h$ is the vertical component of the total displacement of the object, considered positive if upward.

Let's now assume that the object is at rest (in this frame) at times $t_0$ and $t_1$; that's probably implicit in the idea of "lifting". Then the left side of the equation above is zero. So the right side must also be zero. If the forces acting on the object are gravity and only another one, then from the results so far we have that the work done by the other force during this interval in this frame must be $+mgh$.

If we have gravity and two other forces acting on the object, we have that the total work done by the two extra forces together must be $+mgh$. But each force can have done an amount of work different from this. And so on for more forces.

There are many good books about such matters. For example Synge & Griffith's Principles of Mechanics, or Love's Theoretical Mechanics, or Truesdell's A First Course in Rational Continuum Mechanics, which treats all such matters in great depth and with logical care.

Your question has some hidden assumptions. For example I could "lift" the object without applying any force at all – if the object already has an initial upward velocity. Of course you might say "this isn't what I meant by 'lifting'", but this shows the necessity of making questions and statements as clear as possible.

When we speak about "work" we must specify (1) which force is doing the work, (2) in which reference frame the work is measured. In fact two observers in two different frames may assign different works to the same force, because they will measure different velocities. For example the object may be at rest with respect to one observer, so no force is doing any work for that observer. (Note that all observers, inertial and non-inertial, agree on the values of the forces: forces are frame-invariant quantities.)

The work done on a point-like object during a time interval $[t_0,t_1]$ by the force $\pmb{F}(t)$ in a reference frame in which the object has velocity $\pmb{v}(t)$ is $$\int_{t_0}^{t_1} \pmb{F}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$ As the notation indicates, force and velocity may have an arbitrary time dependence. The product $\pmb{F}(t)\cdot \pmb{v}(t)$ is called the power expended by the force $\pmb{F}$ at time $t$ in that reference frame.

Note that the velocity is determined not just by the force $\pmb{F}$, but by all forces acting on the object during the time interval, and by the initial kinematic conditions, such as initial velocity.

According to Newton's laws, the sum of the forces $\pmb{F}^{(1)}, \pmb{F}^{(2)}, \dotsc$ acting on the object must equal, in an inertial frame (such as approximately the one fixed with the floor), the rate of change of momentum in that frame: $$\frac{\mathrm{d} (m \pmb{v})}{\mathrm{d}t} = \sum_{k} \pmb{F}^{(k)}(t)\ .$$

Assuming that the object isn't losing or acquiring mass, if we scalar-multiply this equation by $\pmb{v}$ and time-integrate between $t_0$ and $t_1$ we find $$\tfrac{1}{2}m\, v(t_1)^2 - \tfrac{1}{2}m\, v(t_0)^2 = \sum_k \int_{t_0}^{t_1} \pmb{F}^{(k)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$

For the gravitational force $\pmb{F}^{(\text{grav})} := -m\pmb{g}$ (with $\pmb{g}$ directed upwards) we also find, by simple integration, that $\int_{t_0}^{t_1} \pmb{F}^{(\text{grav})}(t)\cdot \pmb{v}(t)\ \mathrm{d}t = -mgh$, where $h$ is the vertical component of the total displacement of the object, considered positive if upward. We can note again that this displacement is observer-dependent: if we fix a camera on the lifted object, then with respect to the camera the object isn't moving, so $h=0$ in the reference frame of the camera, and the gravitational force hasn't done any work in that frame.

Let's now assume that the object is at rest (in the inertial frame of the floor) at times $t_0$ and $t_1$; that's probably implicit in the idea of "lifting". Then the left side of the equation above is zero. So the right side must also be zero. If the forces acting on the object are gravity and only another one, then from the results so far we have that the work done by the other force during this interval in this frame must be $+mgh$.

If we have gravity and two other forces acting on the object, we have that the total work done by the two extra forces together must be $+mgh$. But each force can have done an amount of work different from this. And so on for more forces.

There are many good books about such matters. For example Synge & Griffith's Principles of Mechanics, or Love's Theoretical Mechanics, or Truesdell's A First Course in Rational Continuum Mechanics, which treats all such matters in great depth and with logical care.

Your question has some hidden assumptions. For example I could "lift" the object without applying any force at all – if the object already has an initial upward velocity. Of course you might say "this isn't what I meant by 'lifting'", but this shows the necessity of making questions and statements as clear as possible.

When we speak about "work" we must specify (1) which force is doing the work, (2) in which reference frame the work is measured. In fact two observers in two different frames may assign different works to the same force, because they will measure different velocities. For example the object may be at rest with respect to one observer, so no force is doing any work for that observer. (Note that all observers, inertial and non-inertial, agree on the values of the forces: forces are frame-invariant quantities.)

The work done on a point-like object during a time interval $[t_0,t_1]$ by the force $\pmb{F}(t)$ in a reference frame in which the object has velocity $\pmb{v}(t)$ is $$\int_{t_0}^{t_1} \pmb{F}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$ As the notation indicates, force and velocity may have an arbitrary time dependence.

Note that the velocity is determined not just by the force $\pmb{F}$, but by all forces acting on the object during the time interval, and by the initial kinematic conditions, such as initial velocity.

According to Newton's laws, the sum of the forces $\pmb{F}^{(1)}, \pmb{F}^{(2)}, \dotsc$ acting on the object must equal, in an inertial frame, the rate of change of momentum in that frame: $$\frac{\mathrm{d} (m \pmb{v})}{\mathrm{d}t} = \sum_{k} \pmb{F}^{(k)}(t)\ .$$

Assuming that the object isn't losing or acquiring mass, if we scalar-multiply this equation by $\pmb{v}$ and time-integrate between $t_0$ and $t_1$ we find $$\tfrac{1}{2}m\, v(t_1)^2 - \tfrac{1}{2}m\, v(t_0)^2 = \sum_k \int_{t_0}^{t_1} \pmb{F}^{(k)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$

For the gravitational force we also find, by simple integration, that $\int_{t_0}^{t_1} \pmb{F}^{(\text{grav})}(t)\cdot \pmb{v}(t)\ \mathrm{d}t = -mgh$, where $h$ is the vertical component of the total displacement of the object, considered positive if upward.

Let's now assume that the object is at rest (in this frame) at times $t_0$ and $t_1$; that's probably implicit in the idea of "lifting". Then the left side of the equation above is zero. So the right side must also be zero. If the forces acting on the object are gravity and only another one, then from the results so far we have that the work done by the other force during this interval in this frame must be $+mgh$.

If we have gravity and two other forces acting on the object, we have that the total work done by the two extra forces together must be $+mgh$. But each force can have done an amount of work different from this. And so on for more forces.

There are many good books about such matters. For example Synge & Griffith's Principles of Mechanics, or Love's Theoretical Mechanics, or Truesdell's A First Course in Rational Continuum Mechanics, which treats all such matters in great depth and with logical care.

Your question has some hidden assumptions. For example I could "lift" the object without applying any force at all – if the object already has an initial upward velocity. Of course you might say "this isn't what I meant by 'lifting'", but this shows the necessity of making questions and statements as clear as possible.

When we speak about "work" we must specify (1) which force is doing the work, (2) in which reference frame the work is measured. In fact two observers in two different frames may assign different works to the same force, because they will measure different velocities. For example the object may be at rest with respect to one observer, so no force is doing any work for that observer. (Note that all observers, inertial and non-inertial, agree on the values of the forces: forces are frame-invariant quantities.)

The work done on a point-like object during a time interval $[t_0,t_1]$ by the force $\pmb{F}(t)$ in a reference frame in which the object has velocity $\pmb{v}(t)$ is $$\int_{t_0}^{t_1} \pmb{F}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$ As the notation indicates, force and velocity may have an arbitrary time dependence. The product $\pmb{F}(t)\cdot \pmb{v}(t)$ is called the power expended by the force $\pmb{F}$ at time $t$ in that reference frame.

Note that the velocity is determined not just by the force $\pmb{F}$, but by all forces acting on the object during the time interval, and by the initial kinematic conditions, such as initial velocity.

According to Newton's laws, the sum of the forces $\pmb{F}^{(1)}, \pmb{F}^{(2)}, \dotsc$ acting on the object must equal, in an inertial frame, the rate of change of momentum in that frame: $$\frac{\mathrm{d} (m \pmb{v})}{\mathrm{d}t} = \sum_{k} \pmb{F}^{(k)}(t)\ .$$

Assuming that the object isn't losing or acquiring mass, if we scalar-multiply this equation by $\pmb{v}$ and time-integrate between $t_0$ and $t_1$ we find $$\tfrac{1}{2}m\, v(t_1)^2 - \tfrac{1}{2}m\, v(t_0)^2 = \sum_k \int_{t_0}^{t_1} \pmb{F}^{(k)}(t)\cdot \pmb{v}(t)\ \mathrm{d}t\ .$$

For the gravitational force $\pmb{F}^{(\text{grav})} := -m\pmb{g}$ (with $\pmb{g}$ directed upwards) we also find, by simple integration, that $\int_{t_0}^{t_1} \pmb{F}^{(\text{grav})}(t)\cdot \pmb{v}(t)\ \mathrm{d}t = -mgh$, where $h$ is the vertical component of the total displacement of the object, considered positive if upward.

Let's now assume that the object is at rest (in this frame) at times $t_0$ and $t_1$; that's probably implicit in the idea of "lifting". Then the left side of the equation above is zero. So the right side must also be zero. If the forces acting on the object are gravity and only another one, then from the results so far we have that the work done by the other force during this interval in this frame must be $+mgh$.

If we have gravity and two other forces acting on the object, we have that the total work done by the two extra forces together must be $+mgh$. But each force can have done an amount of work different from this. And so on for more forces.

There are many good books about such matters. For example Synge & Griffith's Principles of Mechanics, or Love's Theoretical Mechanics, or Truesdell's A First Course in Rational Continuum Mechanics, which treats all such matters in great depth and with logical care.