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A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The spring constant of the spring is 1.00 N/cm. The mass is pulled downward 2.00 cm and released. What is the speed of the mass when it is 1.00 cm above the point from which it was released?

Respuesta :

Chenk13

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

[tex]y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}[/tex]

1. Finding Ф from the initial conditions:

[tex]-0.02 = 0.02cos(0 + \phi) => \phi = \pi[/tex]

2. Finding time t at position y = 1 cm:

[tex]0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}[/tex]

3. Find velocity v at time t from equation 2:

[tex]v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))[/tex]

johanrusli

The speed of the mass is about 0.0866 m/s

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Further explanation

Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.

[tex]\boxed {F = k \times \Delta x}[/tex]

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

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The formula for finding Young's Modulus is as follows:

[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]

E = Young's Modulus ( N/m² )

F = Force ( N )

A = Cross-Sectional Area ( m² )

Δx = Extension ( m )

x = Initial Length ( m )

Let us now tackle the problem !

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Given:

mass of the object = m = 4.00 kg

force constant = k = 1.00 N/cm = 100 N/m

displacement = d = 2.00 cm = 0.02 m

Unknown:

speed of the mass = v = ?

Solution:

Let's find the initial displacement of the spring:

[tex]F = k x[/tex]

[tex]m g = k x[/tex]

[tex]4 \times 9.8 = 100 x[/tex]

[tex]x = 39.2 \div 100[/tex]

[tex]x = 0.392 \texttt{ m}[/tex]

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Next, we will use the law of conservation of energy as follows:

[tex]E_{p1} + E_{k1} = E_{p2} + E_{k2}[/tex]

[tex]\frac{1}{2}k(x + d)^2 + 0 = \frac{1}{2}k(x + 0.01)^2 + mg(0.01) + \frac{1}{2}mv^2[/tex]

[tex]\frac{1}{2}(100)(0.392 + 0.02)^2 = \frac{1}{2}(100)(0.392 + 0.01)^2 + 4(9.8)(0.01) + \frac{1}{2}(4)v^2[/tex]

[tex]8.4872 = 8.0802 + 0.392 + 2v^2[/tex]

[tex]0.015 = 2v^2[/tex]

[tex]v = \sqrt{0.015 \div 2}[/tex]

[tex]v = \frac{1}{20} \sqrt{3} \texttt{ m/s}[/tex]

[tex]v \approx 0.0866 \texttt{ m/s}[/tex]

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Learn more

  • Young's modulus : https://brainly.com/question/6864866
  • Young's modulus for aluminum : https://brainly.com/question/7282579
  • Young's modulus of wire : https://brainly.com/question/9755626

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Answer details

Grade: College

Subject: Physics

Chapter: Elasticity

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Keywords: Elasticity , Diameter , Concrete , Column , Load , Compressed , Stretched , Modulus , Young

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