# Potential Energy (maximum height above initial postion)

1. Nov 20, 2009

### dbzpwns

1. The problem statement, all variables and given/known data

A ball having mass 1.2 kg is connected by
a string of length 1.6 m to a pivot point and
held in place in a vertical position. A constant
wind force of magnitude 23 N blows from left
to right.If the mass is released from the vertical po-
sition, what maximum height above its initial
position will it attain? Assume that the string
does not break in the process. The accelera-
tion of gravity is 9.8 m/s2 .

2. Relevant equations

mgh=PE
1/2at^2=d
Fnet=ma

3. The attempt at a solution
I'm sorry but i couldn't get this problem at all. Could someone please help me with this?

2. Nov 20, 2009

### tiny-tim

Welcome to PF!

Hi dbzpwns! Welcome to PF!

(try using the X2 tag just above the Reply box )

Hint: use the work-energy equation …

work done = change in mechanical energy (W = ∆KE + ∆PE)

3. Nov 20, 2009

### dbzpwns

thank you, tiny-tim for the welcome :D.

so KE = (1/2) (kg) (velocity squared)
and
PE= (gravity) (kg) (meters)

Would that be right? do i add those to get work? after that where should i go?

4. Nov 20, 2009

### tiny-tim

Yes.
No, work done = force "dot" displacement (W = F.x)

5. Nov 20, 2009

### dbzpwns

ok i multiply but how would i find velocity for finding KE?

6. Nov 20, 2009

### tiny-tim

At maximum height, the velocity is … ?

7. Nov 20, 2009

### dbzpwns

im sorry but i dont understand

I think i might have found it. Is it 3.70?

Last edited: Nov 20, 2009
8. Nov 20, 2009

### dbzpwns

hey is everything alright. i was hoping to figure this out ASAP its ok with you man

9. Nov 21, 2009

### tiny-tim

Hi dbzpwns!

(just got up :zzz: …)
uhh?

At maximum height (assuming it doesn't "loop-the-loop" ), the velocity must be zero, mustn't it?

10. Nov 21, 2009

### dbzpwns

yeah i thinking that but i guess i got confused. So would my work be 432.768 Joules? Now, where should i go

11. Nov 21, 2009

### tiny-tim

uhh?? where did 432.768 come from?

the work done will depend on θ (or on height).

12. Nov 21, 2009

### dbzpwns

well you said, work= force x displacement.

so i got 18.816 for PE. it was 0 for KE. i added those and multiplied 23 to get it. If thats wrong than what should i do?

13. Nov 21, 2009

### tiny-tim

You've multiplied the horizontal force (23) by PE.

Start again, call the maximum angle θ …

what is the work done ?

14. Nov 21, 2009

### dbzpwns

i dont know. wouldn't i need a force to find the work

15. Nov 21, 2009

### tiny-tim

?? It's 23N.

16. Nov 21, 2009

### dbzpwns

ok so now im searching for an angle. What should i do with the 23N. I'm sorry but im not really understanding

17. Nov 21, 2009

### dbzpwns

didn't you say that the addition of KE and PE x force would get you work. I did that and you said it was the wrong force

18. Nov 21, 2009

### tiny-tim

Hi dbzpwns!

(is this revision, or is this the first time you've seen this? …)

work done is force times the distance in the direction of the force that the point of application of the force moves.

For example, the work done by gravity is mg (the force) times the difference in height (because the force is vertical, so only vertical distance is relevant … the work done is the same so long as the height is the same, even if there is also horizontal displacement).

In this case, the force (23N) is horizontal, so only the horizontal part of the displacement is relevant.

In other words, if the displacement is d = (x,y,z), and the force is 23î where î is a horizontal unit vector, then the work done is 23(d.î) = 23x.

To put it in simple English, how far does the mass move sideways, and multiply that by 23.

19. Nov 21, 2009

### tiny-tim

No, I said that the addition of KE and PE would be equal to the work done (which equals force "dot" displacement) …

20. Nov 21, 2009

### dbzpwns

oh i understand you. So where would i go from here though, do i multiply kg times 23 (force)