When we lift an object work done by gravitational force is negative $-mgh$ means gravitational force is taking energy from the body. How? Which energy is being reduced by the gravitational force?
[Physics] Work done by gravitational force is negative
newtonian-gravitynewtonian-mechanicspotential energywork
Related Solutions
Lets place the stationary mass $M$ a the origin. In both cases the gravitational force is towards the origin, and is given by $$\vec{F}_{g}=-\frac{GMm}{r^{2}}\hat{r}$$
Thus in the first case, the work done by the gravitational force is $$W_{r\rightarrow\infty}=\int_{r}^{\infty}\vec{F}_{g}\cdot\vec{dr}=-\int_{r}^{\infty}\frac{GMm}{r^{2}}dr=-\frac{GMm}{r}$$
On the other hand, in the second case we just need to sweep the limits of integration $$W_{\infty\rightarrow r}=\int_{\infty}^{r}\vec{F}_{g}\cdot\vec{dr}=-\int_{\infty}^{r}\frac{GMm}{r^{2}}dr=\frac{GMm}{r}$$
which is exactly the opposite, and the signs are just what you've anticipated. An important thing to note is that also in both cases $$\vec{dr}=dr\hat{r}$$
and the sign is set by the integration limits. So in the first case $dr>0$ and in the second $dr<0$ (in a very informal manner). You ignored this fact when you wrote $\cos\left(180^{\circ}\right)$ in the dot product in the second case.
There's always confusion with this topic when it's not well explained. It's all inside "work-energy theorem", which says
$$\Delta E_k = W$$
But we'll make a distinction here: work done by conservative forces and work done by non conservative forces:
$$ \Delta E_k = W_C + W_{NC} $$
And now, we just call "minus potential energy" to the work done by conservative ones
$$W_{C}:= -\Delta E_p$$
We do this for convenience. We can do it, because a conservative force is such taht can be written as a substraction of a certain function $B$ like this:
$$W_C=B(\vec{x_f})-B(\vec{x_0}) $$
We just decide to define $E_p=-B$, so $W_{C}=-\Delta E_p$. We include that minus sign so that we can take it to the LHS:
$$ \Delta E_k = W_C + W_{NC} $$ $$ \Delta E_k = -\Delta E_p + W_{NC} $$ $$ \Delta E_k + \Delta E_p = W_{NC} $$ $$ \Delta E_m = W_{NC} $$
So the increment in mechanical energy is always equal to the work done by non-conservative forces. If there are no non-conservative forces, then $\Delta E_m=0$ and energy is conserved (that's why we call them like that.
(read it slowly and understand it well)
So, having this in mind, I think your confusion arises because of that famous "artificial" negative sign.
There are many formulas, and it's typicall to have a mess. It's all about surnames: $\Delta E_k = W_{Total}$, but $\Delta E_m=W_{NC}$. The subindices are the key.
The force of engines is non-conservative. Hence, their work contributes to total mechanical energy.
Gravity is conservative, so we can work with its potential energy.
If there is no increase of kinetic energy, that means
$0 + \Delta E_p = W_{NC}$
So engines are only increasing potential energy. But that means
$$-W_C = W_{NC}$$
Of course, if there's no gain in KE, no acceleration, there's equilibrium. The work of the engines is compensating the work of gravity.
- Negative work is always positive $\Delta E_p$, by definition.
- More altitude means more $E_p$, you are right. But here energy is not conserved (engines). Normally, increasing height would decrease $E_k$, but we're adding work so taht $E_k$ stays constant.
- $\Delta E_k=0$ implies $W_{Total}=0$. That means gravity is making negative work, and engines are doing positive work (equilibrium). The thing is that potential energy variation is minus gravity's work.
Best Answer
This answer will be a little long but I think you’ll understand by the end of it.
Throw a ball up in the air. You have imparted initial Kinetic Energy (KE) right at the beginning. Observe the ball move up.
As the ball moves up, you see the velocity of the ball is reducing since the force of gravity is acting against it. In Physics, we say that this force of gravity is doing negative work on the ball.
The ball has now reached the top, its velocity is zero. Basically force of gravity has done enough negative work to reduce the velocity to zero and therefore its KE has also becomes zero. Let’s say this total work done by gravity in upward journey is W1 (it would be a negative sign e.g. -4J or -10J)
But what has happened to the initial KE. The KE of the ball keeps reducing as it moves up but another form of energy keeps increasing. This other form of energy is Potential Energy (PE). Thus the PE at start is zero and keeps increasing till all KE has converted to PE by the time it reaches the top of the flight. The gravitational force that did negative work on the ball and decreased its KE has in the process increased the PE of the ball. Thus negative work (W1) has resulted in positive change in PE. According to work KE theorem,
$$\Delta KE = W1 ———- Eq. 1$$
but since Mechanical energy has to be conserved
$$\Delta PE + \Delta KE = 0 ———- Eq. 2$$
Use Eq. 1 to substitute Delta KE as W1 in equation 2, we get
$$\Delta PE + W1 = 0$$
or
$$\Delta PE = - W1$$
As an example, if work done is say -10 J and the change in PE is from say PE (initial) = 0 J to PE (final) = 10 J, then-
$$\Delta PE = -W$$
or $$PE (final) - PE (initial) = -W$$
or $$10 J - 0 J = - (-10 J)$$
$$10 J = 10J$$