Question

A 30.0-g object connected to a spring with a force constant of 30.0 N/m oscillates with an amplitude of 7.00 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 1.05 cm. (Let 0 cm be the position of equilibrium.) m/s.
(c) Find the kinetic energy when its position is 3.50 cm. mJ.
(d) Find the potential energy when its position is 3.50 cm.

Answer 1

Answer:

(a)  E = 73.5mJ

(b) v = 2.188 m/s

(c) KE = 55.1 mJ

(d) PE = 18.37mJ

Explanation:

m = mass of the object = 0.03 kg

k = spring constant = 30 N/m

A = amplitude = 0.07 m

For part (a)

The total energy of the system can be found by:

E = k × A² / 2  

E = (30 N/m) × (0.07 m)² / 2

E = 0.0735 J

E = 73.5mJ

For part (b)

The elastic potential energy in the spring at position d = 0.0105 m is:

PE = k × d² / 2

PE = (30 N/m) × (0.0105 m)² / 2

PE = 0.00165375 J

The kinetic energy at that point is:

KE = E - PE

KE = (0.0735 - 0.00165375)

KE = 0.07184 J

So its speed is:

KE = m × v² / 2

v = √( 2 × KE / m )

v = √( 2 × (0.07184 J) / (0.03 kg) )

v = 2.188 m/s

For part (d)

The elastic potential energy in the spring at position d = 0.035 m is:  

PE = k × d² / 2

PE = (30 N/m) × (0.035 m)² / 2

PE = 0.018375 J

PE = 18.37mJ

For part(c)

The kinetic energy at that point is:

KE = E - PE

KE = (0.0735J) - (0.018375 J)

KE = 0.0551 J

KE = 55.1 mJ