A 0.20-kg block rests on a frictionless,level surface and is attached to a horizontally aligned spring with a spring constant of 80 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?

Question

Quizplus · Accepted Answer

Using conservation of mechanical energy, we can find the maximum speed of the block at the equilibrium point. The potential energy stored in the spring when the block is displaced 4.0 cm is (1/2)(k)(x^2) = (1/2)(80)(0.04) = 1.6 J. At the equilibrium point, all of the potential energy is converted to kinetic energy, so (1/2)(m)(v^2) = 1.6 J. Solving for v, we get v = sqrt((2*1.6)/0.20) = 1.8 m/s. However, we must account for the frictional force which acts in the opposite direction of motion. The work done by friction is F*d = (0.3)(0.04) = 0.012 J. This reduces the amount of mechanical energy available, so the kinetic energy at the equilibrium point is (1.6 - 0.012) J. Solving for v again gives v = sqrt((2*1.588)/0.20) = 0.72 m/s, which is answer choice C.

A 020-Kg Block Rests on a Frictionless,level Surface and Is Attached