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Conservation of energy

  • Thread starter the whizz
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  • #1
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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.
 

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Answers and Replies

  • #2
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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.
 
  • #3
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Here's his problem:

6.) A sphere of mass [tex]m[/tex] hangs from a string of length [tex]l[/tex] and is drawn back so that the string makes an angle of [tex]\theta[/tex] with the vertical axis. It swings down until the strings hits a peg which is [tex]l/2[/tex] 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 [tex]\alpha[/tex]. Use conservation of energy to find an equation for the angle [tex]\alpha[/tex]. Show that this equation makes sense with a few, well-chosen examples for [tex]\theta[/tex].
 
Last edited:
  • #4
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sent. ty.
 
  • #5
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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:

[tex]U_0+K_0=U+K[/tex], where [tex]U[/tex] indicates potential energy and [tex]K[/tex] represents kinetic energy. What do the two critical moments have for their energy?
 
  • #6
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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.
 
  • #7
152
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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 [tex]0m[/tex] is given by: [tex]h=\frac{k}{2mg}d^2-d[/tex]. [tex]k=898N\cdot m[/tex], mass is [tex]0.0907kg[/tex], and [tex]d=27in[/tex]; find [tex]h[/tex]. And, find [tex]v[/tex] at [tex]h=0[/tex].
 
  • #8
32
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Neglect air resistance.

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

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
 
  • #9
152
0
  • #10
32
0
Neglect air resistance.

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

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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