- #1

- 19

- 0

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.

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.