What is the Conservation of Energy Homework Problem?

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Homework Help Overview

The discussion revolves around the conservation of energy in the context of a physics problem involving potential and kinetic energy. The original poster presents a scenario involving an arrow shot upwards, with specific parameters such as spring constant, mass, and distance. Participants are exploring the implications of energy conservation in this setup.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between potential and kinetic energy at different points in the motion of the arrow. There are attempts to derive equations for height and velocity based on given parameters. Some participants question the setup of the equations and the assumptions made regarding energy states.

Discussion Status

There is an ongoing exploration of the problem with various interpretations being discussed. Some participants have provided equations and attempted solutions, while others are seeking clarification on the mathematical setup and the physical principles involved. No consensus has been reached yet.

Contextual Notes

Participants note the importance of neglecting air resistance in their calculations. There is also mention of the challenges in sharing equations through attachments, which may affect the clarity of the discussion.

the whizz
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Homework Statement



attached

Homework Equations



equation in the problem

k = 898 N/m
m = 90.7g = .0907kg
d = 27 inch = .6558m
g = 9.8 m/s

The Attempt at a Solution



h = [(898n/m)/(2)(.0907kg)(9.8m/s)](.6558m)^2 - (.6558)

that is for the H. when computing this is got 217.55m. I am assuming the way the equation is written out for h in the problem that is how to set it up.

now it asks for velocity.

when it gets back to zero. We will say E1 = E2...

at E1 it is all potential energy and no kinetic. and at E2 where we find final velocity it is all kenetic and no potential?



so then we have mgh = 1/2 mv^2...

plus the top equation in for h

then isolate V and solve.

thanks for any help.
 

Attachments

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just curious do most people not like attachments...

Its the best way for me to post the problems because a lot of equations have trouble making readable just typing in.
 
Here's his problem:

6.) A sphere of mass m hangs from a string of length l and is drawn back so that the string makes an angle of \theta with the vertical axis. It swings down until the strings hits a peg which is l/2 directly below the point where the string is attached to the ceiling. When the string hits the peg, the sphere keeps moving and swings up through an angle \alpha. Use conservation of energy to find an equation for the angle \alpha. Show that this equation makes sense with a few, well-chosen examples for \theta.
 
Last edited:
sent. ty.
 
the whizz said:
sent. ty.

The problem's posted in the above post; if you want, you can "quote" me and select what I typed and put it in the original post. Otherwise, here's my suggestion:

You know the conservation of energy equation:

U_0+K_0=U+K, where U indicates potential energy and K represents kinetic energy. What do the two critical moments have for their energy?
 
yes i am sorry that one is the wrong question for the work i included...resent you the right one. I will most likely be able to use that one too though so thank you.
 
the whizz said:
yes i am sorry that one is the wrong question for the work i included...resent you the right one. I will most likely be able to use that one too though so thank you.

Neglect air resistance.

The height of an arrow shot directly upwards from an inital height 0m is given by: h=\frac{k}{2mg}d^2-d. k=898N\cdot m, mass is 0.0907kg, and d=27in; find h. And, find v at h=0.
 
asleight said:
Neglect air resistance.

The height of an arrow shot directly upwards from an inital height 0m is given by: h=\frac{k}{2mg}d^2-d. k=898N\cdot m, mass is 0.0907kg, and d=27in; find h. And, find v at h=0.

the whizz said:

Homework Statement



attached

Homework Equations



equation in the problem

k = 898 N/m
m = 90.7g = .0907kg
d = 27 inch = .6558m
g = 9.8 m/s

The Attempt at a Solution



h = [(898n/m)/(2)(.0907kg)(9.8m/s)](.6558m)^2 - (.6558)

that is for the H. when computing this is got 217.55m. I am assuming the way the equation is written out for h in the problem that is how to set it up.

now it asks for velocity.

when it gets back to zero. We will say E1 = E2...

at E1 it is all potential energy and no kinetic. and at E2 where we find final velocity it is all kenetic and no potential?



so then we have mgh = 1/2 mv^2...

plus the top equation in for h

then isolate V and solve.

thanks for any help.

ok put the two together
 
the whizz said:
ok put the two together

That sounds perfect.
 
  • #10
asleight said:
Neglect air resistance.

The height of an arrow shot directly upwards from an inital height 0m is given by: h=\frac{k}{2mg}d^2-d. k=898N\cdot m, mass is 0.0907kg, and d=27in; find h. And, find v at h=0.

the whizz said:

Homework Statement




k = 898 N/m
m = 90.7g = .0907kg
d = 27 inch = .6558m
g = 9.8 m/s

The Attempt at a Solution



h = [(898n/m)/(2)(.0907kg)(9.8m/s)](.6558m)^2 - (.6558)

that is for the H. when computing this is got 217.55m. I just wanted to check on the math here.

we shoulddo k/2mg first then multiply that times d^2 then subtract d. any check on the solution would be great.
 

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