Why $g$ is not negative in potential energy $= mgh$ formula? For eg. A boy of mass 55 kg runs up a staircase of 50 steps in 10s. If the height of each step is 10 cm, find his power.
My question is here why are we not taking g as negative as boy is moving up against gravity. 
 A: $g$ is a symbol which stands for the magnitude of the free-fall acceleration, which is approximately equal to $9.8$ m/s$^2$ on the Earth's surface.
If you're working in a coordinate system in which $\hat y$ points upward to the sky, then the acceleration of a freely falling object is $\vec a = -g \hat y = -\left(9.8\frac{\text{m}}{\text{s}^2}\right) \ \hat y$.  
Vector quantities (such as position, velocity, acceleration, and force) don't have signs attached to them; they are not positive or negative.  Only the components of a vector relative to some coordinate system can be positive or negative.  For example, the acceleration $\vec a$ of a freely falling object is a vector.  The $y-$component of this vector (in a coordinate system in which $\hat y$ points upward) is negative, and is equal to $-g = -9.8\frac{\text{m}}{\text{s}^2}$.
On the other hand, if we use a coordinate system in which $\hat y$ points downward, then the $y-$component of the acceleration vector is positive, and is equal to $g = 9.8\frac{\text{m}}{\text{s}^2}$.

There are some authors who use the (in my mind) ridiculous convention that $g = -9.8 \frac{\text{m}}{\text{s}^2}$, and is therefore an intrinisically negative quantity.  This is madness for two reasons - firstly, because it leads to formulae such as $v_f = \sqrt{-2gh}$ which looks wrong until you remember the intrinsic minus sign in $g$, but more importantly because it obscures the fact that you still need to define the actual acceleration relative to a coordinate system.
A: It's by definition. I'm not quite sure what your background is, so I'm trying to explain this more in laymen's terms.
I guess your confusion stems from the fact that you know the gravitational force as F = -mg, where g is a positive constant (usually 9.81 m/s²), but here in the potential there is no minus sign.
It comes from the definition of a potential. In physics, you can find the force corresponding to a potential by taking minus the gradient (something like the derivative) of the potential. In your example, the potential is P(x) = mgx, and taking minus the derivative with respect to x gives us the force F(x)= -mg. This has the nice property that our potential energy is positive, but the resulting force is negative and thus attracting.
Increasing your height is supposed to increase your potential energy. If you define the potential with a negative sign, then an increase in height would lead to a decrease in potential energy.
A: I know I'm very late to this question, but I thought I would add some maths to explain it for anyone that's interested. 
As every other answer the sign of $g$ depends on which coordinate system you are in. Normally, we choose the coordinate system to have it's $x, y$ axes parallel the ground and the $z$ axis perpendicular to those and pointing away from the ground, that is, pointing towards the sky. Since the gravitational force acts to "pull" down a particle its direction is opposite to the $z$ axis and therefore we say that $g$ is negative, that is $F_z = -mg$ at the Earth's surface.
To answer your question about why the gravitational potential $U = mgh$ is non negative, it comes down to the way we define a potential. In physics, we usually define the potential (if it exists) as the function $U$ which satisfies 
$$\vec F = -\text{grad} \ U = - \begin{pmatrix} 
\frac{\partial U}{\partial x} \\ 
\frac{\partial U}{\partial y} \\
\frac{\partial U}{\partial z}
\end{pmatrix}$$
where $\vec F$ is a force (notice the negative sign). In this case, since $g$ only affects a particle in the $z$ direction with the magnitude of $-mg$, assuming the $z$ axis points towards the sky, we can write the gravitational force $\vec F$ as the vector field $\vec F = (0 \ \ 0 \ \ -mg)$. In order for $mgh$ to be a potential for this force, its gradient needs to equal $\vec F$, let's check that it indeed is a potetnial. Remember that $h$ represents a distance in the $z$ direction, hence we have for each component 
$$
-\frac{\partial U}{\partial x} = -\frac{\partial}{\partial x} mgh =  0 \\
-\frac{\partial U}{\partial y} = -\frac{\partial}{\partial y} mgh =  0 \\
-\frac{\partial U}{\partial z} = -\frac{\partial}{\partial z} mgh =  -mg 
$$
which clearly shows that $-\text{grad}( mgh )= \vec F$. 
Now, as to why we have chosen the force to equal to the negative gradient, I would say  it's really arbitrary, that is, there is no inherent mathematical reason for it to be negative. But it does however make intuitive sense that $mgh $ is non negative, as it represents the amount of work the gravitational force can do; When we move something higher up the more it can fall, and falling is precisely the work done by the gravitational force on an object. 
In your example, the boy increases his height from some $h_0$ to $h = h_0$ + 50 steps, which gives you a positive change in potential as the gravitational force now needs to be able to pull him down from those 50 steps as well. Let's say he went down 50 steps instead, giving us a negative potential, what would that mean? It means that the gravitational force has done work and now has less ability to work -- pull you down. 

We can also derive the potential given the force $\vec F = (0 \ \ 0 \ \ -mg)$. An equivalent formulation of the definition of potential above is 
$$U = -\int_\gamma \vec{F} \cdot d\vec r.$$
Let $\vec r (t) = (x(t) \ \ y(t) \ \ z(t)), t \in [\alpha, \beta]$ describe the path $\gamma$ of the particle such that $\vec r (\alpha) = \vec a$ and $\vec r (\beta) = \vec b$, where $\vec a, \vec b$ are the start and end point of the path respectively. Then, by the definition of the line integral
$$U = -\int_\gamma \vec F (\vec r)\cdot \vec r' dt = -\int_\alpha^\beta (0 \ \ 0 \ \ -mg) \cdot (x(t)' \ \ y(t)' \ \ z(t)')dt = mg\int_\alpha^\beta z(t)'dt $$
and by the fundemental theorem of calculus we get 
$$ 
U = mgz(\beta) - mgz(\alpha) = mg\left( z(\beta) - z(\alpha) \right)
$$
but $z(\beta) - z(\alpha)$ is just a change in height, let's denote this with $h$ and we get the expected result
$$
U = mgh.
$$
A: You could. Here, +/– represent direction. Ultimately, from situation to situation you can define which direction is positive and negative. If you define up as positive, you can treat the force of gravity as negative.
Bare in mind that energy is a scalar — it has no direction. So for a simple problem like this (with no competing forces, etc.) you can just work with absolute values.
A: I get a lot of students who really, really want the gravitational acceleration, $g$, to have a negative value.
It doesn't. It's a vector. The magnitude of the vector is positive. Its direction is down.
Now, there are times when it is convenient to set up a coordinate system where you represent "down" using negative numbers, and "up" using positive numbers.  But there are other times where that's not convenient.  You have encountered one of those times.
When you find yourself in doubt as to the sign of a quantity in physics, the right way to respond is to use English words (or words in your favorite natural language) to describe what is happening.  Here you have a boy who is climbing stairs.  Once he reaches the top, he'll have more gravitational potential energy --- that is, energy stored in the gravitational field between the boy and the Earth --- than he did before.  Energy enters the gravitational field at a rate of 275 watts, as you have computed already.  That energy is coming from chemical reactions in the boy's muscles, which must be giving off energy at a rate somewhat larger than 275 watts, because some of the energy from the chemical reactions is wasted as heat and noise.
Now for energy-balance problems, it will be important to make sure that the side which is losing energy and the side which is gaining energy are represented with opposite algebraic signs. But if all you're doing is finding the magnitude of the change, then the sign is not relevant.  Consider a problem where a car is going up a ramp, rather than a boy up the stairs.  I think that since energy is leaving the car's engine and getting converted into gravitational potential energy, it makes sense to talk about the power of the engine as negative.  But in all the marketing literature I've ever read about cars, I've never seen one advertised with negative horsepower.  It's not a convenient convention in that case.
As an aside, a fifty-step climb in ten seconds is some pretty impressive footwork.
