Question

A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of 39.5 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless

Answer 1

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

Answer 2

Answer:

The speed of the mass is 7.74 m/s

Explanation:

Given data:

m = 35.9 g = 0.0359 kg

k = 18.4 N/m

A = 39.5 cm = 0.395 m

The displacement in spring is:

$x=\frac{A}{2} =\frac{0.395}{2} =0.1975m$

The law of conservation of energy:

initial energy = final energy

$\frac{1}{2} kA^{2} =\frac{1}{2} kx^{2} +\frac{1}{2} mv^{2} \\kA^{2}=kx^{2}+mv^{2}\\(18.4*0.395^{2} )=(18.4*0.1975^{2} )+(0.0359v^{2} )\\v=7.74m/s$