Question

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?

Answer 1

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.

$\boxed {F = k \times \Delta x}$

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:

$\boxed {E = \frac{F / A}{\Delta x / x_o}}$

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:

$F = k x$

$m g = k x$

$4 \times 9.8 = 100 x$

$x = 39.2 \div 100$

$x = 0.392 \texttt{ m}$

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

$E_{p1} + E_{k1} = E_{p2} + E_{k2}$

$\frac{1}{2}k(x + d)^2 + 0 = \frac{1}{2}k(x + 0.01)^2 + mg(0.01) + \frac{1}{2}mv^2$

$\frac{1}{2}(100)(0.392 + 0.02)^2 = \frac{1}{2}(100)(0.392 + 0.01)^2 + 4(9.8)(0.01) + \frac{1}{2}(4)v^2$

$8.4872 = 8.0802 + 0.392 + 2v^2$

$0.015 = 2v^2$

$v = \sqrt{0.015 \div 2}$

$v = \frac{1}{20} \sqrt{3} \texttt{ m/s}$

$v \approx 0.0866 \texttt{ m/s}$

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Learn more

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

Answer 2

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

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

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

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

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

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

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$