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 7.2 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 0.079 m to the right of this position. The block is pulled to the right, stretching the spring by 0.055 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

For the block to hit the bottle,

the elastic potential energy of the spring at the bottle(x = 0.079m) should be equal to the sum of the elastic potential energy of the spring at x = 0.055m and the kinetic energy of block at x = 0.055m

Now the potential energy of the block at x = 0.079m is

where k is the spring constant given by

ω is the angular frequency of the oscillation and m is the mass of the block.

Therefore potential energy of the spring at the bottle(x = 0.079m) is

  

now the elastic potential energy of the spring at x = 0.055m is

and the kinetic energy of the block at x = 0.055m is

then we should have

  

Answer 2

the block is oscilating with 7.2 rad /s = ω ;

conserevtion of energy :

ΔK + ΔP = 0 ;

m/2 ( 0 ^2 - v0^2 ) + k /2 [ 0.079^2 - 0.055^2 ] = 0 ;

m/2 ( 0 ^2 - v0^2 ) + [ ω^2 m ]/2 [ 0.079^2 - 0.055^2 ] = 0 ;

1/2 ( 0 ^2 - v0^2 ) + [ ω^2 ]/2 [ 0.079^2 - 0.055^2 ] = 0 ;

1/2 ( 0 ^2 - v0^2 ) + [ 7.2^2 ]/2 [ 0.079^2 - 0.055^2 ] = 0 ;

v0 = 0.4083104 m /s <---------ans