A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Energy work calculations (please check over)

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