How are energy and work related?

[teggenspiller]

how are energy and work related??

Homework Statement

so there is potential and kinetic energy, which together or separate make up mechanical energy. pe comes in two types, spring(elastic) and gravitational. potential energy=mass*grav*height. kinetic energy applies to objects who are moving. ke+1/2*m*v^2

okay, so that's all fine and dandy, i get that.
next, i have work. Work=force*displacement*cos(theta)-->angle between the force and the displacement vector.

but what happens when i have a question that asks me something such as
7. A 5.0-kg brick is moving horizontally at 6.0 m/s. In order to change its speed to 10.0 m/s, the net work done on the brick must be:

A. 40 J
B. 90 J
C. 160 J
D. 400 J
E. 550 J
Right Points Earned: 1/1
Your C

i infact, got it correct.
using knet=kfin-kin

however, the info they give me
m=5
v=6

vf=10
is good for finding the KINETIC ENERGY (refer to equations)

but they specifically ask for "net WORK"

i got it right by doing the only thing that made since, finding the net kinetic energy...


(2. Homework Equations )

so does this must mean the KE is closely related to WORK...

(3. The Attempt at a Solution )

but how?

 

[Doc Al]

so does this must mean the KE is closely related to WORK...

Most definitely! See: http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr"

 

[thegreenlaser]

Take a look at the equation you used:

$$K_{net} = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}m (v_f^2 - v_i^2) = \frac{1}{2}m(v_f - v_i)(v_f + v_i) = \frac{(v_f + v_i)}{2} (v_f - v_i) m$$
$$K_{net} = (v_{average})(\Delta v) m$$

Note that:
$$v_{avg} = \frac{\Delta d}{\Delta t}$$
and
$$a = \frac{\Delta v}{\Delta t} \Rightarrow \Delta v = a \Delta t$$

So,
$$K_{net} = (v_{average})(\Delta v) m = \frac{\Delta d}{\Delta t}(a \Delta t)m = (\Delta d)(ma) = F \Delta d$$
$$K_{net} = W$$

It may not be immediately obvious, but the equation you used is simply a derivation of the work formula, assuming that force applied is the same as the direction of motion, so that the work formula is just $$W = F \Delta d cos(0) = F \Delta d$$. In this problem everything is in one line, so the displacement, velocity, acceleration, and force vectors are all parallel, so that is a fair assumption.

 

[teggenspiller]

whooa!