Spring compression and velocity with conservation of energy

[the whizz]

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.

 

[the whizz]

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.

 

[asleight]

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 $$\theta$$.

 

[the whizz]

sent. ty.

 

[asleight]

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?

 

[the whizz]

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.

 

[asleight]

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 $$h=0$$.

 

[the whizz]

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 $$h=0$$.

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

 

[asleight]

ok put the two together

That sounds perfect.

 

[the whizz]

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 $$h=0$$.

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.

 

[the whizz]

http://img253.imageshack.us/img253/6248/question5xu1.th.jpg http://g.imageshack.us/thpix.php

original problem

 

[the whizz]

my solution for the first part.

http://img372.imageshack.us/img372/2766/problem5answerkd2.th.jpg http://g.imageshack.us/thpix.php