Questions on Energy changes on a ball

  • Thread starter adam640
  • Start date
  • #1
19
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Hi, this is a question from a past paper, for which I have the exam on Monday. Please see the attachments.

Solutions
a) w = fx
=(20)(1.5-0.7)
=16J

b)PE=mgh
^PE=(.25)(10)(1.5)-(.25)(10)(.7)
PE=2J

c) ?
when i attempted a solution i used ^KE=^PE
this gave me an answer of v^2=30
I don't understand how I find the initial speed using only energy arguments?
Any help would be greatly appreciated.
 

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Answers and Replies

  • #2
318
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You know how much energy you put into the ball from a). You know how much potential energy the ball has from b).
Where is the rest of the energy?
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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welcome to pf!

hi adam! welcome to pf! :smile:

(have a delta: ∆ and try using the X2 icon just above the Reply box :wink:)
c) … when i attempted a solution i used ^KE=^PE

do you mean the work energy theorem ∆KE + ∆PE = work done ?

show us your calculations :smile:
 
  • #4
6
0
Remember that energy has been put into the system (the ball is moved upward by a hand), so just equating PE and KE will not work, an additional term describing the energy added to the system must be incorporated.

As Betel said you know how much energy has been added to the system (from part (a) of the question)

Good luck with the exam!
 
  • #5
19
0
Thanks!

KE = 1/2 mv2
(16-2) (2) = (0.25)v2
v = (sqrt)112ms-1
 

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