jack1234 said:
Thanks a lot, I think this is the crux I don't understand, and I am very eager to get it clear :)
If we take downward as positive why it can't be mg(+x)?
We can choose the directions for position... and we can choose the 0 energy position... but after that, gravitational potential energy is fully determined... we can't choose beyond this point...
we can choose the positive direction and negative direction for position... but after that we can't choose when it comes to energy... we can choose a reference energy and that's it.
suppose we take downwards as positive. x>0.
and suppose we choose energy at x = 0, to be 0.
now gravitational potential energy at all other points is full determined.
so what is the gravitational potential energy at x=x1, where x1>0
by definition of potential energy... it is the negative of the work done by gravity in moving from x = 0 to x = x1.
gpe at x =x1 is:
-\int_0^{x_1}{\vec{F}\cdot\vec{dx}
=-\int_0^{x_1}{mg*dx} (notice that mg acts downwards so it is positive.)
= -mgx1
suppose instead we took downwards as negative and upwards as positive... we want the gravitational potential energy at the same position... but then the same position is located at x = -x1 .
now we need gpe at x = -x1
-\int_0^{-x_1}{\vec{F}\cdot\vec{dx}
=-\int_0^{-x_1}{-mg*dx} (now mg is negative since downwards is negative here.)
= \int_0^{-x_1}{mg*dx}
= -mgx1
so what's the main point here... no matter what coordinate system we use... the gravitational potential energy at a point below the 0 energy position will be negative. Same way no matter what coordinate system we use, the gravitational potential energy at a point above the 0 energy position will be positive.
suppose we have some 0 position x = 0... now suppose we take up positive down negative... we have some position x =x1.
what is the gravitational potential energy at x=x1...
it is:
0 + negative of work done by gravity to go from 0 to x1
0 + [-(-mg)(x1-0)] = mgx1
if we switch to up negative downpositive. now we want potential energy at a point x = x1 (using the new coordinate system).
0 + negative of work done by gravity to go from 0 to x1
0 + [-mg(x1-0)] = -mgx1
but remember... these give the same results... because a position of x =x1 in the old coordinates has coordinates of x = -x1 in the new coordinates... leading to the same gravitational potential energy!
If we take downward as negative, why it is not k(-x)(-x)?
In everything below, I'm assuming that the mass hangs below the 0 position...
if we take down as negative... and suppose x>0... that means that the position is actually -x in this coordinate system... spring force = -k*displacement (this is always true)
spring force + gravitational force = 0
-k(-x-0) - mg = 0
kx = mg.
what is the total energy... -mgx + (1/2)kx^2 = -(1/2)kx^2
suppose we take up as negative... down as positive... and again x is given as >0... this time the position is x in this coordinate system.
-k(x-0) + mg = 0
kx = mg
what is the total energy... -mgx + (1/2)kx^2 = -(1/2)kx^2
Now... suppose x<0 as it is given... (ie they're giving a negative number). suppose we take down as negative up as positive... we can leave x as it is because it fits with the coordinate system...
-k(x-0) - mg = 0
-kx = mg
what is the total energy at this x value... mgx + (1/2)kx^2 = -kx^2 + (1/2)kx^2 = -(1/2)kx^2
suppose x<0 as it is given. suppose we take positive down negative up... we need to switch to -x because it is below 0 (we assume this from the context of the problem)...
-k(-x-0) + mg = 0
-kx = mg
what is the total energy at this x value... mgx + (1/2)kx^2 = -kx^2 + (1/2)kx^2 = -(1/2)kx^2
lol! Looks like I rambled on a lot... hope what I wrote makes sense.
A main point... the x could be positive or negative... but as long as the actual position is below the 0 energy position... we get a result of -(1/2)kx^2.
