Question

Chapter 10, Problem 81 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 8.4 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 Vo Ignoring the width of the block, find vo i 0.050 m 0.080 m Number Units the tolerance is +/-2%

Answer 1

In order for the block to knock over the bottle, it must have a potential energy of

Ep = ½kx² = ½k(0.08m)² = 0.0032m²·k

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

So, 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) ---------------------------------(i)

As given in the problem, ω = 8.4 rad/s
and we know that ω = sqrt(k/M);

put this value in (i), we have -

v = 0.06m * w = 0.06 * 8.4 = 0.504 m/s