Work Energy Theorem

[Heat]

1. The problem statement, all variables and given/known data

You throw a rock of weight 20.4 N vertically into the air from ground level. You observe that when it is a height 14.9 m above the ground, it is traveling at a speed of 26.0 m/s upward.

Use the work-energy theorem to find its speed just as it left the ground;
Use the work-energy theorem to find its maximum height.
Take the free fall acceleration to be g = 9.80 m/s^2.


2. Relevant equations
Wtotal = delta K

K = mv^2

v^2 = vi^2 + 2a(x-xi)

3. The attempt at a solution


I believe this can also be done using kinematic equation, since we are given vf,x,a, and we want vi.

So if this is true, I will attempt both ways :D


26^2 = vi^2 + 2(-9.8)(14.9-0)
vi = 31.11 <--------> which is right o:)

now, for using the work energy theorem.

we first need the mass.

using
w=mg

20.4 = m ( 9.8)

m = 2.08

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2

K2 = 843.65
K1 = 0

....how do I proceed?

[stunner5000pt]

since there is no external force (such as friction or air resistance) there is no external work done on theb all except for gravity. So the total initial energy of the ball is equal to the total final energy of the ball

what is the initial energy of the ball and the final energy of the ball?

[Heat]

ummm...

so Winitial = Wfinal.

W=Fs

W = (F)(14.9)...?

[nrqed]

ummm...

so Winitial = Wfinal.

W=Fs

W = (F)(14.9)...?

No, it's not "work initial" = "work final", which in itself would not make sense.

you may either use

change of kinetic energy = total work done by gravity

OR

total final energy = total initial energy

where total energy = kinetic energy plus gravitational potential energy

[Heat]

delta k = wg ?

if so, how do I apply this to my problem, as I have this:

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2

[stunner5000pt]

delta k = wg ?

if so, how do I apply this to my problem, as I have this:

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2
ok so u have the final and initial kinetic energies
now how do you calculate the work done by gravity??

[Heat]

w done by gravity is:

w = mgsin(theta)

w = 2.08 (9.8)(sin 90)
w = 20.38