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 of 8.3rad/s. The drawing shows the position of the block when the springis unstrained. This position is labeled ''x = 0 m.'' Thedrawing also shows a small bottle located 0.080 m to the right ofthis position. The block is pulled to the right, stretching thespring by 0.050 m, and is then thrown to the left. In order for theblock to knock over the bottle,it must be thrown with a speedexceeding v0. Ignoring the width of the block,find v0.

Explanation / Answer

   for this we use the law of conservation ofmechanical energy which is the final total mechanical    energy Ef is equal to the initialtotal mechanical energy Ei    from the theory the total mechanical energyis given as    E = (traslational kinetic energy) +    (rotational kinetic energy) +    (gravitational potential energy) +    (elastic potential energy)    E = (1 / 2) m v2 + (1/ 2) I 2 + m g h + (1 / 2) k x2    so as Ef = Ei we canwrite    (1/2) m vf2 +(1/2) I f2 + m g hf + (1/2)k xf2 = (1/2) mvo2 + (1/2) Ii2 + m g hi + (1/2) kxi2    according to uor question    f =i        = 0    as the block reaches the bottle thefinal velocity will be zero 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 bothsides and cancelling (1/2) we get    vo2 = (k / m)(xf2 - xi2)    vo = (k / m)(xf2 - xi2)    the angular frequency of simple harmonicmotion is given by     = (k / m)    so we get    vo = xf2 - xo2         = .............m /s         = .............m /s

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