Question

A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and b) the maximum speed of the object if the amplitude of the motion is 3.00 cm. (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm

Answer 1

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J