Chapter 16, Problem 77PQ

A lightweight spring with spring constant k = 225 N/m is attached to a block of mass m₁ = 4.50 kg on a frictionless, horizontal table. The block–spring system is initially in the equilibrium configuration. A second block of mass m₂ = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m₂ loses contact with the spring–block 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B).

a. What is the subsequent speed of block 2?

b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a).

77. (a) The energy of the system initially is entirely potential energy.

$E_0=U_0=\frac{1}{2}k{y}_{\max }^2=\frac{1}{2}(225\text{ N/m)(0}{\text{.150 m)}}^2=2.53\text{ J}$

At the equilibrium position, the total energy is the total kinetic energy of both blocks:

$\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(4.50\text{ kg}+3.00\text{ kg)}{v}^2=(3.75\text{ kg})v^2=2.53\text{ J}$

Therefore, the speed of each block is

$v=\sqrt{\frac{2.53\text{ J}}{3.75\text{ kg}}}=\boxed{0.822\text{ m/s}}$

(b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or $v=-0.822\text{ m/s}$.

FIGURE P16.77