Question

I'm trying to write a paper for my physics class, and this is what my teacher is asking me to include:

Answer 1

Derive the work-energy theorem starting with \[a = F/m\] and \[Vf^2 = Vi^2 +2ad\] and ending with \[W=DeltaKE\]

Answer 2

Imagine throwing a ball vertically in a gravitational field. The initial velocity is Vi and the final velocity is Vf and during that time interval t, the ball travels a distance d.

Using the kinematic equation you provide to describe the motion, the acceleration, a, is g.
Using Newton's second law you provided, F=ma...The only force here is gravity F=-mg

If you recall from earlier topics, you conclude from kinematics that the height the ball reaches is found to be:
height = d = Vi^2/(2*g) -----rearranging to get
Vi^2/2 = g*d -----multiply both sides by m
1/2(m*Vi^2) = m*g*d

recall: Work = Force *distance = (-)m*g*d
KE = 1/2(m*V^2)

1/2(m*Vi^2) = m*g*d
Work = KE

In this example, the height the ball travels in the gravity field is d, and when the ball reaches the top, the KE is zero.

Work = Delta(KE)

Answer 3

\[Wa \rightarrow b =\int\limits_{a}^{b} F*dx\]

\[F=m*a=m*\frac{ dv }{ dt } \]

\[dx = v*dt\]

\[Wa \rightarrow b = \int\limits_{a}^{b}m*\frac{ dv }{ dt }*v*dt = \int\limits_{a}^{b}m*v*dv \]

\[Wa \rightarrow b = m*1/2*Vb ^{2} - m*1/2*Va ^{2} \]