When to use the work-energy formula

[StephenDoty]

I am having trouble deciding when to use the work-energy formula; change in KE=W and derivations of this formula, or the Kf+Uf+Eth=Ki+Ui+Wext formula when presented with work problems.

Are there any type of guidelines or hints that anyone could give me so I can more easily decide which formula to use?

Thank you.
Stephen

 

[Defennder]

What is Eth and Wext supposed to be here?

 

[StephenDoty]

Eth= thermal energy
Wext= work done by external surfaces

 

[Defennder]

What do "external surfaces" refer to? You meant instead to say "external forces" correct? As a general rule you have look out for keywords in the question to know whether or not you have to take heat loss into account. I'll start with the most inclusive formula, the one you quoted and then read the question to see which factors can be neglected.

 

[Redbelly98]

I am having trouble deciding when to use the work-energy formula; change in KE=W and derivations of this formula, or the Kf+Uf+Eth=Ki+Ui+Wext formula when presented with work problems.

Are there any type of guidelines or hints that anyone could give me so I can more easily decide which formula to use?

Thank you.
Stephen

You can always use
Kf+Uf+Eth = Ki+Ui+Wext

With some problems, this becomes
Kf = Wext
because all the other terms are zero.

With some other problems, it becomes
Kf + Uf = Ki + Ui
because Eth and Wext are zero.

You have to read the problem carefully and figure out which terms are zero (if any), then you simplify the equation whenever possible.

 

[StephenDoty]

I was talking in general but an example would be:
A 5kg box is attached to one end of a spring with a spring constant of 80N/m. The other end of the spring is attached to a wall. Initially the box is at rest at the spring's equillibrium position. A rope with a constant tension of 100N then pulls the box away from the wall. What is the speed of the box after it has moved .5m? The coefficient of friction between the box and floor is .3.

My professor solved this problem using the formula: Kf + Uf + Delta Eth = Ki + Ui + Wext.
Wext is the work of the external forces which would be the rope.
Wext = Tcos theta * Delta r.
Wext = Tcos0 * .5m = T*0.5m = (100N)*.5m = 50J

To find the Delta Eth We have to find Wdiss.
Wdiss = f_k *cos theta * Delta r = f_k * cos180 * .5m = -f_k * .5m = -$$\mu$$mg*.5m= -(.3*5kg*9.8m/s/s * .5m)= -7.35J

And since Delta Eth= - Wdiss, Delta Eth= 7.35J

Now we can plug data into Kf + Uf + Delta Eth = Ki + Ui + Wext.
1/2mvf^2 + 1/2kxf^2 + 7.35J = 1/2mvi^2 + 1/2kxi^2 + 50J

vi=0m/s xi= 0m
thus
1/2mvf^2 + 1/2kxf^2 + 7.35J = 50J
vf=$$\sqrt{[2(50J-7.35J-(1/2kxf^2)]/m}$$
thus
vf= 3.61m/s

This was the way my professor did this problem, but couldn't we use the energy-work formula : Delta K=Wc + Wdiss + Wext
Wdiss = -Delta Eth

1/2mvf^2 - 1/2mvi^2 = -Delta U +-7.35J + 50J
where the 50J is the Wdiss and the 7.35J is the Wext that we found earlier.
1/2mvf^2 - 1/2mvi^2 = (Ui-Uf) + 50J - 7.35J
And since vi=0m/s, xi=0m and U=1/2kx^2
1/2mvf^2-0=(1/2k(0)^2)-(1/2kxf^2) +42.65J
1/2(5kg)vf^2=-(1/2(80N/m)(.5m)^2) + 42.65J
5/2kg * vf^2= -10J + 42.65J
2.5kg *vf^2 = 32.65J
vf^2= 13.06m^2/s^2
vf=3.61 m/s

Which is the same answer that my professor got. So, is there any reason why I cannot use the energy-work formula, Delta K=Wc + Wdiss + Wext
, and W= F*Delta r * cos theta, instead of Kf + Uf + Delta Eth = Ki + Ui + Wext?

Thank you for all of your replies.
Stephen

 

[Doc Al]

Assuming that you have the required information, you can use either conservation of energy or the "work-energy" theorem. (Realize that the latter really comes from Newton's 2nd law, analyzing the forces acting on the body, not conservation of energy.)