Question

A mass of 2.5 kg is attached to a spring that has a spring constant of 65 N/m. The mass is oscillating on a frictionless, horizontal surface. When the spring is stretched 10 cm, its elastic potential energy and the kinetic energy of the mass are equal. What is the maximum speed of the mass?

Answer 1

Answer 2

Answer 3

when spring is stretched by 0.1m PE of string = 1/2 k x² = 0.5*65 * (0.1)² =0.325 J

TE = PE+ KE = 2*PE = 0.65J

for (KE)_max, PE=0 => KE=TE

                            => 0.5*m*v² =0.65

                             => v² = 1.3/2.5=0.52

                             => v= 0.72 m/s

Answer 4