Question

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 8.9 rad/s. The drawing shows 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 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 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.

Answer 1

It sounds like this is the setup:
/\/\/\/\/\/\/\?<---- 0.080m ---->?
pull block 0.050m right, throw left, oscillates once and knocks over bottle.
Sounds interesting.
To just knock over the bottle, the spring must attain a potential energy of
Ep = ½kx² = ½k(0.080m)² = 0.0032m²·k

At x=0.05m, the spring has a potential energy of
Ep = ½k(0.05m)² = 0.00125m²·k

Therefore, it needs to be given a kinetic energy of
Ek = (0.0032m² - 0.00125m²)·k = 0.00195m²·k = ½Mv²
v = sqrt(0.00195m²·2·k/M) = 0.06m·sqrt(k/M)

But we're given that ? = 9.1 rad/s
and we know that ? = sqrt(k/M); so
v = 0.06m·8.9rad/s = 0.534 m/s