Question

A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. (a) Find the total energy of the system. mJ (b) Find the speed of the object when its position is 1.20 cm. (Let 0 cm be the position of equilibrium.) m/s (c) Find the kinetic energy when its position is 2.50 cm. mJ

Answer 1

Answer:

(a) The total energy of the spring system is 0.032 J

(b) The speed of the object when its position is 1.20 cm is approximately 1.28996 m/s

(c) The kinetic energy when its position is 2.50 cm is 0.0195 J

Explanation:

The given parameters are;

The mass of the object connected to the spring, m = 35.0 g = 0.00

The force constant, k = 40.0 N/m

The amplitude of the oscillation, a = 4.00 cm = 0.04 m

Therefore, we have

(a) The total energy of the spring system, E given as follows;

E = PE + KE = 1/2·m·v² + 1/2·k·x²

Where;

v = The velocity of the spring

x = The extension of the spring

When the spring is completely extended, x = a, and v = 0, therefore;

The total energy of the spring system, E = 1/2 × k × a² = 1/2 × 40.0 N/m × (0.04 m)² = 0.032 J

(b) At x = 1.20 cm = 0.012 m, we have;

E = 1/2·m·v² + 1/2·k·x²

0.032 = 1/2 × 0.035  × v² + 1/2 ×  40 × 0.012²

0.032 - 1/2 ×  40 × 0.012² = 1/2 × 0.035  × v²

0.02912 = 1/2 × 0.035  × v²

1/2 × 0.035  × v² = 0.02912

v² = 0.02912/(1/2 × 0.035) = 1.664

v = √1.664 ≈ 1.28996

The speed of the object when its position is 1.20 cm,  v ≈ 1.28996 m/s

(c) When its position is 2.50 cm = 0.025 m, we have;

E = PE + KE

0.032 = 1/2 ×  40 × 0.025² + KE

KE = 0.032 - 1/2 ×  40 × 0.025² = 0.0195

The kinetic energy when its position is 2.50 cm = 0.0195 J.