A 0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of 40 N/m.The block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion.A frictional force of 0.3 N exists between the block and surface.What is the speed of the block when it passes through the equilibrium point after being released from the 4.0-cm displacement point?
A)0.45 m/s
B)0.63 m/s
C)0.80 m/s
D)1.2 m/s

Question

Quizplus · Accepted Answer

Explanation:The work done by the frictional force is:$$W_f = F_f d = (0.3 N)(0.04 m) = 0.012 J$$The initial potential energy of the spring is:$$U_i = \frac{1}{2}kx_i^2 = \frac{1}{2}(40 N/m)(0.04 m)^2 = 0.032 J$$The final kinetic energy of the block is:$$K_f = U_i - W_f = 0.032 J - 0.012 J = 0.02 J$$The speed of the block is:$$v = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(0.02 J)}{0.20 kg}} = 0.45 m/s$$

A 020-Kg Block Rests on a Frictionless Level Surface and Is