# Find greatest acceleration and speed

1. May 16, 2013

### songoku

1. The problem statement, all variables and given/known data
A winch is used to raise a 200 kg load. The maximum power of the winch is 5 kW. Calculate the greatest possible acceleration of the load when its speed is 2 ms-1, and the greatest speed at which the load can be raised.

2. Relevant equations
W = F.d
W = ΔKE
P = F.v
P = W/t
KE = 1/2 mv2

3. The attempt at a solution
Assume the initial speed = 0 m/s:
W = ΔKE = 0.5 x 200 x 4 = 400 J

Then do not know what to do.....

The answer should be 0.2 ms-2 and 2.55 ms-1

Thanks

2. May 16, 2013

### tiny-tim

hi songoku!
use power = energy per time = force times distance per time = force times speed

(and newton's second law)

3. May 16, 2013

### Simon Bridge

P=Fv
Draw a free body diagram to find F.

4. May 16, 2013

### songoku

I still don't get it. What I did:

F = P/v = 5000 / 2 = 2500 N

ƩF = m.a
2500 - 2000 = 200 . a
a = 2.5 ms-2

My logic: there are 2 forces acting on the object, one is from the winch directed upwards and the other is weight. By using F = P/v, I find the force by the winch then put it in second law of motion.

5. May 16, 2013

### tiny-tim

yes that looks fine

i got the same as you … the book answer seems to be wrong

(but i get the right answer for the second part! )

6. May 16, 2013

### songoku

hi tiny-tim

How to get the answer for the second part?

Using formula v = P/F, to get the greatest speed means that power has to be the greatest and the force should be minimum. Greatest power is 5000 W and minimum force is 2000 N, so v = 2.5 ms-1?

Thanks

7. May 17, 2013

### Simon Bridge

The first answer wants the maximum acceleration, the second wants the maximum speed.
It seems convenient to do them in reverse order.

I wouldn't normally do a homework problem, but it should be safe here since OP has already done the work and has pretty much understood the material.

from the fbd:
$T-mg=ma$ ( where T=tension in rope, and "up" is positive - assumes no losses)
... so

$T=m(a+g)$ ...(1)

power from the winch must be $P=Tv$ ...(2)

For constant speed, T=mg so the power needed is P=mgv

 oops - I stand corrected!
Hopefully the walk-through didn't go through yet

@songoku: for the second section - the velocity is constant, so what is the acceleration?
Therefore what is the tension in the rope.

Last edited: May 17, 2013
8. May 17, 2013

### songoku

Hope I don't misinterpret your work.

By putting T = m(a+g) to P = Tv:
P = m(a+g)v
a = P/mv - g = 2.5 ms-2?

Constant velocity means zero acceleration so the tension will be the same as weight (2000 N). Then v = P/F = 5000/2000 = 2.5 ms-1?

Both answers are the same as I have posted above.

Thanks

9. May 17, 2013

### Simon Bridge

You seem to be using g=10N/kg - the book answer suggests they expect you to use g=9.8N/kg.

... and that is what I answered.
Your reasoning appeared indirect, and you seemed uncertain, so I figured you'd benefit from a more direct path.

To get the book's a=0.2m/s/s requires P/mv = 10, what does the speed have to be?
How does that compare to the maximum speed?
Is that sensible?

You could try figuring out what values of P and m would give you the book values.

10. May 17, 2013

### songoku

Oh ok. I get it. So let me redo it a little:
To find greatest speed: v = P / F = 5000 / 1960 = 2.55 ms-1

The speed has to be 2.5 ms-1 to obtain acceleration 0.2 ms-2 by using a = P/mv - g

I think it is still sensible if we compare it with maximum speed, but the question asks when the speed is 2 ms-1 so I think the greatest acceleration at that speed can't be 0.2 ms-2. Do I get it right?

Thanks

11. May 17, 2013

### Simon Bridge

Well done!
You've just disproved the book answer for the initial assumptions.
It [the book] is either, wrong, or, the assumptions about the problem were wrong.

It looks to me like the problem got edited just before publication and the answers were not.
You may find it is fixed in the next edition.

The power of science/math is such that a novice can overturn an established authority.
If you know other people working through the same problems, you should contact them and compare.

12. May 17, 2013

### songoku

Hahaha I like the way you said it.

Thanks a lot for the help, simon and tiny-tim