- #1

blackout85

- 28

- 1

1/2mv= 1/2KX^2

max PE equals starting KE.

then follows

x= v/sqrt k

x=.25 , that is the answer I am getting. But how is the book getting .05m

my second question:

A 6.0kg block is released from rest 80m above the ground. When it has fallen 60m its kinetic energy is:

I got 1200J as my answer

mgh=(6.0*9.81* 60m)=3531.6

PE + KE= (ME + KE)

3531.6 + KE= (6.0*9.81*80m)

KE= 1177.2

is that right