danvalen1 said:
Using 9.8 m/s2:
w=(1/2)(F)(d)
As Filip Larsen points out, you are not using the correct formula for the work done by gravity.
I'm using w=(1/2)Fd because it is derived from w=(1/2)kd2 since k*d = F.
The formula for the force on a spring,
F = kd
is correct.
The formula for the potential energy of a spring, or work done on a spring,
W = \frac{1}{2}kd^2
is also correct. Let me show you where this comes from. By the definition of,
W = \int \vec F \cdot d \vec s
we have as applied to a spring (using the variable
x instead of
d, for the compression distance),
W = \int _0 ^x \vec F(x') \cdot d \vec x'
= \int _0 ^x kx dx,
Thus
W = \frac{1}{2} k x^2.
Or if you use d for the compression distance,
W = \frac{1}{2} k d^2.
But,
W \ne \frac{1}{2}Fd.
It doesn't "work" that way.
For gravity, assuming a constant acceleration, thus constant force over h,
W = \int \vec F \cdot d \vec s
becomes,
W = \int _0 ^h \vec F \cdot d \vec z
= \int _0 ^h mg dz.
But here (unlike the spring case), m and g (and F, for that matter) are constant over z, and can be pulled out from under the integral.
W = mg \int _0 ^h dz
Which becomes,
W = mgh
Or if you'd rather,
W = F _g h
where F
g is the force due to gravity (assuming constant acceleration).
The major difference between gravity and a spring, is that in the case of gravity, the force is constant. In the case of a spring, the force is proportional to the amount of compression. So they each have correspondingly different equations for their work functions (or potential energy functions, if you interpret it that way).
