# Homework Help: Work done by engine pulling up a mass

1. Jan 18, 2012

### jiboom

just a query about last part of a question.

a mass m hangs at end of light string and is raised vertically by an engine working at a constant rate kmg.at t=o,velocity =0 x=0 and x is measured in upwards direction.

now i have shown increase in total energy of the mass, when moved up a height h, due to this motion is

m(k^2lnB-ku)

it nows wants to deduce the time taken by considering the work done by the engine.

now i have said

work done is ∫kmg dx between 0 and h
so let x=t then dx=dt and x=o,t=o x=h t=t
then work done is kmgt

energy at bottom is 0 so energy gained by mass=work done by engine

so

m(k^2lnB-ku)=kmgt

ie gt=(klnB-u)

now im sure this cant be right reasoning, but the final answer is correct.

firstly,is my expression for work done by engine correct?

secondly is my energy consideration correct? it says in my book in one place work done by force is change in ke and derives this then says later work done is equal to the change in mechanical energy,so which is the case?

the book even does a question of a mass going down a plane and calculates the velocity by working out work done by force displacing particle and equating this to KE,no mention of PE?

very confused

2. Jan 18, 2012

### Delphi51

In your last example, the mass ends up at height zero, so there is not PE to include.

I may be misunderstanding your question. I don't know what u and B are for starters.
In lifting the mass with force kmg to height h, the work done is W = F*d = kmgh.
No integral needed since the force is constant.

To work out the time taken, I would use F = ma, a = F/m = kmg/m = kg.
This is constant, so we can use the constant accelerated motion formula
d = ½at² or t² = 2d/a = 2h/kg

3. Jan 21, 2012

### jiboom

delphi51:

sorry,but vague with U,b as they are not important,i have worked out the values earlier on in the question.

what i am asking is if what i have done is correct. can i take the integral of work and transform to time as i habve done and equate this to total energy gain?

with regards to the last example,there is PE,it may end at bottom but it started with PE mgh so total energy at top is mgh+mv^2/2 at bottom it was mu^2/2

but the solution said work done was mv^2/-mu^2/2?

4. Jan 21, 2012

### Delphi51

Sorry; looks like I misunderstood the question. What is B in your answer? What is k? You said the "engine was working at constant rate kmg". I took k to be a constant with no units;and kmg to be a force.
Here you use kmg as a force.

In your answer you have that mku is an energy. Therefore k must have units of velocity. Then the "rate" kmg would have units of power, not force.

No use continuing until we know what kmg is.

5. Jan 22, 2012

### jiboom

no, you are correct. the question says kmg is a constant rate so k is constant. u is the initial velocity.

i know kmg is constant but the chapter i was working on was about w=∫F dx

so i just guessed for ∫kmg dx the F=1 not kmg,so im looking at

kmg ∫ 1 dx

then i transform to dt to get work done in in terms of time.

i expect this is what is rubbish???

6. Jan 22, 2012

### Delphi51

If k has no units and kmg is a force, then the work done is W = F*d = kmgh.
Not m(k^2lnB-ku). In fact, this answer doesn't have the units of energy and its two terms have inconsistent units.

"an engine working at a constant rate kmg" does suggest kmg is a power.
If so, it would do work W = P*t = kmgt while pushing the mass up to height h.
That work would go into both potential and kinetic energy so kmgt = mgh + ½mv².
We don't know t or v so something more is needed. Perhaps something could be done with the motion to find another relationship between time and height. I don't see how to do it, though.

Can you ask for clarification on what kmg is?

7. Jan 22, 2012

### jiboom

that is all the question says. a mass is raised by an engine working at a constant rate kmg.

i can fill in the rest of the question but will look messy with no latex.

the equation of motion of the mass is

v^2[dv/dx]=(k-v)g

where v is the upward velocity, x is the displacement measured upwards.

i have mislead you here a bit, i thought u was initial velocity, BUT

initially mass is at rest and when it has risen to a height h its speed is u. then i show

gh=k^2ln[k/(k-u)]-ku-u^2/2

( i have written k/(k-u) as b in my posts above)

then i have shown increase in the total energy of the MASS due to this motion is

m[k^2ln[k/(k-u)]-ku]

and im left to show,by considering the work done by the engine that the time taken is

1/g{kln[k/(k-u)]-u}

which leads to my attempt at transforming a work integral from dx to dt so i have time then equating to total energy increase. it gives the correct answer,but is any of it the correct thing to do??

8. Jan 23, 2012

### Delphi51

I have no idea how you got that equation!

Now you have k with units of velocity. That does make sense; then kmg is a power according to the formula P = F*v. But k is a constant; implying that the mass is in motion at constant speed and the problem much simpler. At constant speed, the force of the engine would just cancel out the mg of gravity, the work done mgh, the time t = d/v= h/k.

It might be helpful if you provided the exact wording of the question.

9. Jan 23, 2012

### jiboom

the question is as written in my post.

i got that equation by :

we have power=kmg=driving forcexvelocity
so driving force =kmg/v upwards

mg acting down

f=ma =>

kmg/v-mg=ma

g(k-v)=av

but a=vdv/dx so

v^2dv/dx=g(k-v)

10. Jan 23, 2012

### jiboom

i think i have seen the answer to my question in another forum,relating to a another question.

engine works at kmg for t seconds so work done is kmgt

then work done by engine=gain in energy

which is basically what i was asking:

is work done just power x time?

is work done =change in energy?

if so why does my book say work done is just change in ke?

11. Jan 23, 2012

### Delphi51

Neat, I haven't seen a = v*dv/dx for years. I see it comes from a related rates chain rule.

Are you sure we have ALL the information from the question? The part you wrote actually doesn't even have a question . . .

What did you think of the idea that it is constant speed? In that case your
v^2dv/dx=g(k-v)
simply says k = v.

12. Jan 23, 2012

### Delphi51

is work done just power x time?
Yes, if power is constant.

is work done =change in energy?
Yes.
if so why does my book say work done is just change in ke?
That would only be for the case when there is no change in other types of energy.

13. Jan 24, 2012

### jiboom

delphi51: thanks for your patience wioth this. i think i now see where my integrating method goes wrong and how to correct. I see now when you said i was using kmg as a force i was completely wrong.

i need

work= ∫F.dx

but i have

kmg=F.v

v=dx/dt

∫kmg dt= ∫F.dx

work=kmgt :)

14. Jan 24, 2012

### jiboom

as for the change in energy query: they derive work doNE is change in ke which i can pass off as it is motion in a line so no PE. but they then do the following question:

a particel of mass m slides from rest down a plane inclined at 30 degrees to the horizontal, the resisitance to the motion of the particle is ms^2 where s is the displacement of the particle from its initial position. find the velocity of the particle when s=1

their solution is:

force down plane is F=mg/2-ms^2

then if v is velocity at s=1,by considering the work done by F in displacing the particle 1 unit down from initial position

∫ F ds {between 0 and 1} =mv^2/2-0

if PE =o is initial position then when its gone down slope 1 unit it has lost mgcos60 PE ??

15. Jan 24, 2012

### Delphi51

The PE is taken care of through the mg/2 term in the force. In integrating it over the distance 0 to 1, it amounts to mg*1*cos(60).

I had work = kmgt ages ago, directly from the definition Power = work/time, but didn't realize that's what you were looking for. I thought t was unknown and the energy had to be expressed in terms of the givens.

16. Jan 25, 2012

### jiboom

im not sure i understand this.

i would have

work done = ∫ F ds {between 0 and 1}

= mg/2-m/3

energy at top=0 pe+ 0 ke=0

energy after 1 unit

mv^2/2-mg/2

so

change in energy is

mv^2/2-mg/2

hence work done =change in energy gives

mv^2/2-mg/2=mg/2-m/3

BUT the book has

mv^2/2=mg/2-m/3

so if the force accounts for the PE,how would i know that? i wish applied maths books explained things better. It does not mention if mg is in the force i dont consider change in potential energy. all it says is work done is change in energy and work is the integral of the force.

Last edited: Jan 25, 2012
17. Jan 25, 2012

### Delphi51

Suppose there was no force dependent on a. Just F = mg/2, component of weight along the ramp. Then the work done as it goes down would be -mgs/2.
The initial PE was mgh = mgs*sin(30) = mgs/2. After going down distance s, its PE is 0.
The initial mgs/2 energy was converted into kinetic energy through the acceleration caused by the mg/2 force.

In the problem, the initial mgh = mgs/2 is partly lost through the friction force ms².
W = integral (-ms²)ds = - ms³/3
so there is a little less KE at the bottom, mgs/2 - ms³/3.
[I find it difficult to replace s with 1 because that makes it look like the dimensions aren't right]

With your calculus skill, you can probably find the velocity after distance s from the acceleration caused by the forces and do a complete check on the book's mv^2/2=mg/2-m/3 without using energy.

18. Jan 25, 2012

### jiboom

i think i get what you mean.

i mistaking the F=ma with the F needed for W=f.d

to do energy method, i would only need to integrate -ms^2 to find the loss of energy due to the work done

then i would have

-ms^3/3=change in energy=mv^2/2-mg/2

:)

now why the book could not just do it tlike this insted of confusing me into counting the work done by gravity twice....

i think you have solved all my troubles with this now. i cant thank you enough.

if its not been to painful for you i will post another thread with some more questions in a while. i have different answers from the book so just need a check on some and a couple im stumped.

19. Jan 25, 2012

### Delphi51

It was a pleasure; thanks for staying with the problem.