# Conservation of energy

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.

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.

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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 $$\theta$$.

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sent. ty.

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.

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 $$h=0$$.

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 $$h=0$$.

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.

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

ok put the two together
That sounds perfect.

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.