Recent content by dbzpwns
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Potential Energy (maximum height above initial postion)
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- dbzpwns
- Post #28
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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...- dbzpwns
- Post #26
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
are you just going to ignore me :(- dbzpwns
- Post #24
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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...- dbzpwns
- Post #23
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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...- dbzpwns
- Post #21
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
oh i understand you. So where would i go from here though, do i multiply kg times 23 (force)- dbzpwns
- Post #20
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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- dbzpwns
- Post #17
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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
- Post #16
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
i don't know. wouldn't i need a force to find the work- dbzpwns
- Post #14
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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?- dbzpwns
- Post #12
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
yeah i thinking that but i guess i got confused. So would my work be 432.768 Joules? Now, where should i go- dbzpwns
- Post #10
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
hey is everything alright. i was hoping to figure this out ASAP its ok with you man- dbzpwns
- Post #8
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
im sorry but i don't understandI think i might have found it. Is it 3.70?- dbzpwns
- Post #7
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
ok i multiply but how would i find velocity for finding KE?- dbzpwns
- Post #5
- Forum: Introductory Physics Homework Help
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Potential Energy (maximum height above initial postion)
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?- dbzpwns
- Post #3
- Forum: Introductory Physics Homework Help