 #1
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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)(1.50.7)
=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.
Solutions
a) w = fx
=(20)(1.50.7)
=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.
Attachments

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