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nikitadnepr [17]

3 years ago

14

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

2 answers:

lidiya [134]3 years ago

8 0

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

stiks02 [169]3 years ago

7 0

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

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

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