Question

A 65.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 5.00 cm on a frictionless, horizontal surface (a) Find the total energy of the system. We know the spring constant and the maximum displacement of the object. What is the value of the spring potential energy and the kinetic energy at that instant? mJ (b) Find the speed of the object when its position is 1.10 cm. (Let 0 cm be the position of equilibrium.) m/s (c) Find the kinetic energy when its position is 2.50 cm. d) Find the potential energy when its position is 2.50 cm. mJ

Answer 1

Here ,

m = 0.065 Kg

k = 40 N/m

A = 0.05 m

a) total energy of the system = 0.50 * k * A^2

total energy of the system = 0.50 * 40 * 0.05^2

total energy of the system = 0.05 J = 50 mJ

total energy of the system is 50 mJ

b) let the speed is v

Using conservation of energy

0.50 * 0.065 * v^2 + 0.50 * 40 * 0.011^2 = 0.05

solving for v

v = 1.21 m/s

the speed of the object is 1.21 m/s

c)

kinetic energy = 0.05 - 0.50 * 40 * 0.025^2

kinetic energy = 0.0375 J = 37.5 mJ

kinetic energy is 37.5 mJ

d)

potential energy = 50 - 37.5

potential energy = 12.5 mJ