Question

A 45.0-g object connected to a spring with a force constant of 50.0 N/m oscillates with an amplitude of 7.00 cm on a frictionless, horizontal surface.

(a) Find the total energy of the system.

(b) Find the speed of the object when its position is 1.30 cm. (Let 0 cm be the position of equilibrium.)

(c) Find the kinetic energy when its position is 3.50 cm.

(d) Find the potential energy when its position is 3.50 cm.

Answer 1

Mass of the object = m = 45 g = 0.045 kg

Spring constant = k = 50 N/m

Amplitude of the oscillation = A = 7 cm = 0.07 m

Total energy of the system = E

E = kA²/2

E = (50)(0.07)²/2

E = 0.1225 J

Position of the object = X₁ = 1.3 cm = 0.013 m

Speed of the object when it is at 1.3 cm = V₁

The total energy of the system is equal to the sum of the potential energy of the spring and the kinetic energy of the object.

E = kX₁²/2 + mV₁²/2

0.1225 = (50)(0.013)²/2 + (0.045)V₁²/2

V₁ = 2.293 m/s

New position of the object = X₂ = 3.5 cm = 0.035 m

Potential energy when the object is at 3.5 cm = PE₂

PE₂ = kX₂²/2

PE₂ = (50)(0.035)²/2

PE₂ = 3.0625 x 10^-2 J

Kinetic energy of the object when it is at 3.5 cm = KE₂

E = PE₂ + KE₂

0.1225 = 3.0625x10^-2 + KE₂

KE₂ = 9.1875 x 10^-2 J

a) Total energy of the system = 0.1225 J

b) Speed of the object when it's position is 1.3 cm = 2.293 m/s

c) Kinetic energy when the object's position is 3.5 cm = 9.1875 x 10^-2 J

d) Potential energy when the object's position is 3.5 cm = 3.0625 x 10^-2 J