• Support PF! Buy your school textbooks, materials and every day products Here!

About potential energies

  • Thread starter jaron
  • Start date
  • #1
23
0

Homework Statement


a wrecking ball, with a mass of 315kg, hangs from a crane on 10.0m of cable. if the crane swings the wrecking ball so that the angle that the cable makes with the vertical is 30', what is the potential energy of the wrecking ball in relation to its lowest position?
What will be the kinetic energy of the wrecking ball when it falls back to the vertical position?
What will be the speed of the wrecking ball?


Homework Equations


i think i will need to find the distance between the ball and the wall sin30'(10m)
Ek = 1/2mv^2
Eg = mgh

the two questions before i am having a hard time with cause i believe i need the final velocity before hitting wall (the speed in this case) before i can solve the others.
but i could be totally wrong (wuldnt be the first time, ;)


The Attempt at a Solution


sin30'(10) = 5m (this is the distance from the ball to the wall.
after that i do: Ek' + Eg' = Ek + Eg
and i am lost because i know there is something i am not doing but dont know what.

Ek' + Eg' = Ek + Eg
1/2mv^2 + mgh = 1/2mv^2 + mgh (mass canceled)
1/2v^2 + (-9.81)(5) = 1/2(0) + 9.81(0) (1/2 canceled)
v^2 = 49.1
v = 7m/s
this is the speed of the wrecking ball that i get.

the answers in my textbook are:
- Eg = 4140 J
- Ek = 4140 J (these two make sense because the two equal one another always
- V = 5.12m/s
---------------------------------------
any help is always appreciated
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
sin30'(10) = 5m (this is the distance from the ball to the wall.
Make a drawing and be a little more careful. That is the horizontal distance. But it is not the vertical distance the ball is above the point of impact. It's the vertical distance that determines the m*g*h
 
  • #3
23
0
i had tried cos 30'(10) = 8.6m
but i figured since the actual movement was the horizontal distance you would assume it to be the height (like a horizontal fall)?

and with that i just end up getting 9.2m/s for my speed. still wrong
 
  • #4
LowlyPion
Homework Helper
3,090
4
i had tried cos 30'(10) = 8.6m
but i figured since the actual movement was the horizontal distance you would assume it to be the height (like a horizontal fall)?

and with that i just end up getting 9.2m/s for my speed. still wrong
If you drew the picture you would see that L*cosθ is the length from the top. But potential energy is how high the bottom of L it is, so what you are interested in is L*(1 - cosθ)
 

Related Threads for: About potential energies

Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
7K
Replies
4
Views
3K
Replies
5
Views
3K
Replies
14
Views
595
Replies
5
Views
689
Top