A block rests on a frictionless horizontal surface and isattached to a spring. W

Question

A block rests on a frictionless horizontal surface and isattached to a spring. When set into simple harmonic motion, theblock oscillates back and forth with an angular frequency of6.5 rad/s. The drawing indicates theposition of the block when the spring is unstrained. This positionis labeled "x = 0 m." The drawing also shows a smallbottle located 0.084 m to the right ofthis position. The block is pulled to the right, stretching thespring by 0.056 m, and is then thrownto the left. In order for the block to knock over the bottle, itmust be thrown with a speed exceeding v0.Ignoring the width of the block, find v0.
m/s Thank you for your help! :) A block rests on a frictionless horizontal surface and isattached to a spring. When set into simple harmonic motion, theblock oscillates back and forth with an angular frequency of6.5 rad/s. The drawing indicates theposition of the block when the spring is unstrained. This positionis labeled "x = 0 m." The drawing also shows a smallbottle located 0.084 m to the right ofthis position. The block is pulled to the right, stretching thespring by 0.056 m, and is then thrownto the left. In order for the block to knock over the bottle, itmust be thrown with a speed exceeding v0.Ignoring the width of the block, find v0.
m/s Thank you for your help! :)

Explanation / Answer

    for this we use the law of conservation ofmechanical energy which is the final total mechanicalenergyEf is equal to the
    initial total mechanical energy

from the theory the total mechanical energy is given as
E = (traslational kinetic energy) +
(rotational kinetic energy) +
(gravitational potential energy) +
(elastic potential energy)
E = (1 / 2) m v2 + (1 / 2) I2 + m g h + (1 / 2) k x2
so as Ef = Ei we can write
(1/2) m vf2 + (1/2) If2 + m g hf + (1/2) kxf2 = (1/2) m vo2 + (1/2) I i2 + m g hi +(1/2) k xi2
according to uor question
f = i
    = 0
as the block reaches the bottle the final velocity will bezero so we can write
vf = 0
and more over
hf = hi
so we get the net form as
(1/2) k xf2 = (1/2) kxi2 + (1/2) m vo2
dividing the above equation with m on both sides and cancelling(1/2) we get
vo2 = (k / m) (xf2 -xi2)
vo = (k / m) (xf2 -xi2)
the angular frequency of simple harmonic motion is given by
= (k / m)
so we get
vo = xf2 -xo2
     = .............m / s

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