Potential Energy (maximum height above initial postion)

[dbzpwns]

Homework Statement

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 .
Answer in units of m.

Homework Equations

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

The Attempt at a Solution

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

 

[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) ​

 

[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?

 

[tiny-tim]

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

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

Would that be right?

Yes.

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

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

 

[dbzpwns]

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

 

[tiny-tim]

… If the mass is released from the vertical po-
sition, what maximum height above its initial
position will it attain?

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

At maximum height, the velocity is … ?

 

[dbzpwns]

im sorry but i don't understandI think i might have found it. Is it 3.70?

 

[dbzpwns]

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

 

[tiny-tim]

Hi dbzpwns!

(just got up :zzz: …)

im sorry but i don't understand


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

uhh?

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

 

[dbzpwns]

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

 

[tiny-tim]

uhh?? where did 432.768 come from?

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

 

[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 that's wrong than what should i do?

 

[tiny-tim]

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 that's wrong than what should i do?

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

Start again, call the maximum angle θ …

what is the work done ? ​

 

[dbzpwns]

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

 

[tiny-tim]

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

?? It's 23N.

 

[dbzpwns]

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

 

[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

 

[tiny-tim]

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

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. ​

 

[tiny-tim]

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

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

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

 

[dbzpwns]

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

 

[dbzpwns]

hey, would you mine just listing everything i have to do to get to the answer. Maybe plug in the numbers too. Give me the equations and what i have to find. I don't mean to sound rude but we both keep logging off and that kind of disrupts my way of thinking on this problem. Please, if you don't mind just tell me how to reach the answer

 

[tiny-tim]

dbzpwns, I've already given you all the equations you need, more than once, and even an extended demonstration of how to do it in the similar case of gravity.

In addition, you should have the benefit of your professor's notes and/or your books.

It's up to you to actually do it.

(Nobody on this forum is going to do it for you.)

If you want to try it again, I'll be happy to comment, and to help.

 

[dbzpwns]

so i find work and that's my answer. No offense man but I am pretty sure that's not even close to the answer since i asked someone else and they said its really long. I'm not telling you to do it for me. I'm asking if you can guide me till the final end and then i plug in the numbers to get the final answer.Those can't be the only equations u needed. I'm still trying man. Please help

 

[dbzpwns]

are you just going to ignore me :(

 

[tiny-tim]

(just got up :zzz: …)

are you just going to ignore me :(

You ignored me first!

No, seriously …

your apparently random attempts are as if you haven't bothered to read any of my hints properly; you didn't answer when i asked …

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

and you haven't taken up my offer …

If you want to try it again, I'll be happy to comment, and to help.

If you read the forum guidelines, you'll see that we don't do people's work for them.
​

 

[dbzpwns]

Again, I am not saying for you to do the work for me. I've never seen this type of problem before. I'm trying to analyze your hints but you never told me how to find the vertical force. I'm sorry but I just want you guide me through the problem. The only reason I am asking is because we both keep logging off and its annoying

 

[tiny-tim]

… you never told me how to find the vertical force. …

vertical force ?

The only vertical forces are gravity, which is taken care of by PE = mgh, and the (partly vertical) tension, which does no work because it's always perpendicular to the motion.

 

[dbzpwns]

so then after i find the potential energy (1.2 X 9.8 X 1.6), which force should i multiply by to get work done. You told me multiplying horizontal force was wrong.

Generally, what should i do after i find the potential energy above

 

[tiny-tim]

You told me multiplying horizontal force was wrong.

Generally, what should i do after i find the potential energy above

No, i said multiplying the PE by the horizontal force was wrong …

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

Don't multiply the PE by anything!

PE is energy. Work done is energy. Leave PE as it is.

Find the work done by the horizontal force (for a particular angle, or for a particular height).

I think I'd better repeat my post #18 …

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. ​

And I'll ask again … have you actually covered work done in your lectures or in your book?