Question

A 0.39-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant) is 570 n/m. the block is pulled from its equilibrium position at x = 0.000 m to a displacement x = 0.080 m and is released from rest. the block then executes simple harmonic motion along the horizontal x-axis. when the position of the block is x = 0.057 m, its kinetic energy is closest to a) 0.90 j. b) 0.84 j. c) 0.95 j. d) 1.0 j. e) 1.1 j

Answer 1

The kinetic energy of a 0.39-kg block on a horizontal frictionless surface, attached to an ideal spring with a spring constant of 570 n/m when the block is at x=0.057m, is closest to 0.84 J, which is option b).

Step-by-step explanation:

The block is initially displaced from the equilibrium position and released, leading to simple harmonic motion. To calculate the kinetic energy of the block when it is at a position of x = 0.057 m, we first need to know the maximum potential energy. The maximum potential energy, which occurs at the maximum displacement (0.080 m), is given by U = (1/2)kx2 = (1/2)(570 N/m)(0.080 m)2. The total mechanical energy of the system is conserved and is equal to the maximum potential energy as there is no friction. When the block is at x = 0.057 m, some of this energy has been converted to kinetic energy. The potential energy at x = 0.057 m is U' = (1/2)kx'2 = (1/2)(570 N/m)(0.057 m)2. The kinetic energy (K) at x = 0.057 m can be found by subtracting the potential energy at that point from the total mechanical energy: K = (Total Mechanical Energy) - U'. When we perform the calculations, we find that the kinetic energy is closest to 0.84 J, which corresponds to option b).