Simple harmonic motion frictionless block problem

[triplel777]

Homework Statement

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 6.6 rad/s. The drawing indicates the position of the block when the spring is unstrained. This position is labeled "x = 0 m." The drawing also shows a small bottle located 0.084 m to the right of this position. The block is pulled to the right, stretching the spring by 0.056 m, and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.

Homework Equations

The Attempt at a Solution

1/2mv_0^2=1/2kd^2
d=0.84-0.56=0.28
V_0=sqrt (k/m)*d
V_0= 6.6*0.28= 1.85

what am i doing wrong?

 

[Delphi51]

1/2mv_0^2=1/2kd^2
d=0.84-0.56=0.28

Something wrong here. You need to say there is KE + spring energy initially and at least spring energy finally. Probably final KE as well since it has to hit fast enough to knock over a bottle. How fast is that?

Subtracting d values is not going to give you the same result as subtracting their squares.

 

[Andrew Mason]

You have to find the energy needed to stretch the spring so that the block reaches the bottle. This means the spring must stretch to .084 m.

After finding the k of the spring (what is the relationship between $\omega$ and k and m?) write out the expression for the energy of the system (block+spring) at the point of maximum stretch when the bottle is struck. Write out the expression for energy of the system when the block is launched. How are the two expressions related? (hint:Assume that energy is conserved).

AM