Amount of work: clarifications

Question

If to lift an object of mass $m$ a force greater than $mg$ is required then why do we take the amount of work done in lifting an object as $mgh$ could someone please explain that to me?

Answer 1

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.

Answer 2

It's not actually. What you say is when at the final point, kinetic energy $\mathcal K$ of the mass is zero. We imagine the experiment is done so gently. If the applied force on the object is greater than $mg$, so it would have acceleration and the speed wouldn't be zero at the end. So the extra energy (extra work done, I mean $F\cdot d-mgh$) is the kinetic energy.

Answer 3

mgh is the work done against gravity to lift the object to height h. If throughout the process you apply a force >mg the object will have some kinetic energy, K, when it is at height h and its total energy will be mgh + K. The work done against gravity is mgh.

If you want the object to be at rest at height h, you first apply an upward force >mg to get the object moving upward. Then as it approaches height h the interplay between your applied force and gravity gets a bit complicated as the force of gravity has to be used to decelerate the object but not so much as to have it start moving downward. So for some period of time you reduce your applied force so there is a net downward force but at the end your applied force equals the force of gravity with the object at rest. So in the integral of force over the distance h, there are periods when your applied force is > mg and periods when it is < mg. Whatever the details of the process, you have done mgh work against gravity.