Thank you for visiting 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. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
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
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Rewritten by : Jeany
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:
[tex]x=\frac{A}{2} =\frac{0.395}{2} =0.1975m[/tex]
The law of conservation of energy:
initial energy = final energy
[tex]\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[/tex]
