Conservation of Energy Question?

Question

A 0.750kg block is firmly attached to a a very light spring (k=275N/m). It is noted that the block-spring system, when compressed 8.0cm and released, stretches out 3.6cm beyond the equilibrium position before stopping and turning back. What is the speed of the block as it passes the equilibrium position for the first time after being released? THE ANSWER IS 1.03m/s, HOW DO YOU GET IT?

Shouldnt we include friction of thermal energy?

Answer 1

You said it: conservation of energy.

Energy before
=
spring potential = 1/2 k(xbefore)^2
=
energy after
=
spring potential = 1/2k(xafter)^2
+
kinetic energy = 1/2 mv^2

They give you k, m, both positions. Solve for v.

This is kind of a wierd problem. Most blocks won't hang on to the spring past equilibrium. But this one apparently does and gets held up somewhat because of it. Anyway, just do the algebra, plug, and chug.

Edit: in real life, yes, you would need to consider energy wasted by non-conservative forces like friction. It textbook problems, they often let you ignore that for simplicity. If they wanted you to consider friction, they would have given you a coefficient of friction or similar information.

Answer 2

PE=.5ks^2
PE=.5(275)(.08)^2
PE=.88
KE=.5mv^2
energy is conserved, so
.88=.5(.75)v^2
v=1.53m/s
this is correct