A block rests on a frictionless horizontal surface and is attached to a spring.
ID: 2253167 • Letter: A
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 8.3 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.
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 8.3 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.Explanation / Answer
k = spring constant
M = mass of the block
natural frequency Wn = (k/M)^0.5 = 8.3 rad /sec
total energy of the system just as the block is thrown = energy in the spring due to stretching + kinetic energy of the block
this energy is converted into the spring energy as the block knocks of the bottle
let v be the velocity with which it is thrown
(0.5 * k * ( 0.05 )^2) + (0.5 * M * v^2) = 0.5 * k * (0.08^2)
dividing both sides by M
0.00125 k/M + 0.5(v^2) = 0.0032 k/M
0.5(v^2) = 0.00195 k/M
v^2 = 0.0039 k/M
v = 0.0624 (k/m)^0.5
v = 0.0624 * 8.3
v = 0.52 m/s
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