Question

A 8.07 kg object on a horizontal frictionless surface is
attached to a spring with *k* = 2180 N/m. The object is
displaced from equilibrium 69.5 cm horizontally and given an
initial velocity of 13.6 m/s back toward the equilibrium position.
What are **(a)** the motion's frequency,
**(b)** the initial potential energy of the
block-spring system, **(c)** the initial kinetic
energy, and **(d)** the motion's
amplitude? **Answers to 3 significant digits (no
scientific notation). Please provide unit of
measurement.**

Answer #1

(a)

T = 2π ∙ √( m / k )

and

f = 1 / T

so

f = 1 / (2π ∙ √( m / k ))

f = √( k / m ) / 2π

f = √( (2180 N/m) / (8.07 kg) ) / 2π

f = **2.62 Hz**
< - - - - - - - - - - - - - - answer (a)

(b)

Ep = k × x² / 2

Ep = (2180 N/m) × (0.695 m)² / 2

Ep = **526.5 J**
< - - - - - - - - - - - - - - answer (b)

(c)

Ek = m × v² / 2

Ek = (8.07 kg) × (13.6 m/s)² / 2

Ek = **746.3 J**
< - - - - - - - - - - - - - - answer (c)

(d)

The total mechanical energy is:

Em = Ep + Ek

Em = (526.5 J) + (746.3 J)

Em = 1272.8 J

Amplitude is the same as maximum stretching or compressing of the spring.

That will occur when all energy is converted to potential energy, so

Ep = k × A² / 2

(1272.8 J) = (2180 N/m) × A² / 2

A = **1.08 m**
< - - - - - - - - - - - - - - answer (d)

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