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!

 

[rwisz]

I can explain the relationship between energy and work by first defining these terms. Energy is the ability of an object to do work, while work is the transfer of energy from one object to another. In simpler terms, energy is the potential to do work, and work is the actual action of using that energy.

In the given question, we can see that the brick has kinetic energy due to its motion. In order to change its speed, a force must be applied to the brick, which will result in the transfer of energy and thus, work being done. This work will change the brick's kinetic energy, allowing it to achieve the desired speed.

The equation for work, W = F * d * cos(theta), shows that work is directly proportional to force and displacement. This means that the more force applied to an object, the more work will be done on it. Similarly, the more distance an object is displaced, the more work will be done on it. In the context of the given question, the force applied to the brick will change its speed, thus changing its kinetic energy and the amount of work done on it.

In conclusion, energy and work are closely related as energy is required to do work, and work is the transfer of energy. In the given question, the change in kinetic energy of the brick is directly related to the work done on it to achieve the desired speed.