Question

A mass m is attached to a spring which is held stretched a distance x by a force f (fig. 7-28), and then released. the spring compresses, pulling the mass. assuming there is no friction, determine the speed of the mass m when the spring returns: (a) to its normal length (x = 0); (b) to half its origin inal extension (x/2).

Answer 1

The speed of the mass m when the spring returns:

(a) to its normal length ( x = 0 ) → v = √ ( F x / m )

(b) to half its original extension ( x/2 ) → v = √ ( 3F x / 4m )

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Further explanation

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension ( compression ) ( m )

Let us now tackle the problem!

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Given:

initial extension of the spring = x

force = F

mass of the object = m

Asked:

speed of the mass = v = ?

Solution:

We will use Conservation of Energy formula to solve this problem.

Part (a) :

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k x^2 + 0 = 0 + \frac{1}{2}m v^2$

$\frac{1}{2}k x^2 = \frac{1}{2}m v^2$

$k x^2 = m v^2$

$v^2 = k x^2 \div m$

$v = \sqrt{ \frac {k x^2}{ m } }$

$\boxed {v = \sqrt{ \frac {F x}{ m }} }$

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Part (b) :

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k x^2 + 0 = \frac{1}{2}k (x/2)^2 + \frac{1}{2}m v^2$

$\frac{1}{2}k x^2 - \frac{1}{2}k (x/2)^2 = \frac{1}{2}m v^2$

$\frac{3}{4}k x^2 = m v^2$

$v^2 = \frac{3}{4} k x^2 \div m$

$v = \sqrt{ \frac {3k x^2}{ 4m } }$

$\boxed {v = \sqrt{ \frac {3F x}{ 4m } }}$

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Learn more

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• Simple Harmonic Motion : brainly.com/question/12069840

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Answer details

Grade: High School

Subject: Physics

Chapter: Elasticity

Answer 2

Spring is stretched by force f to distance "x"

now here by force balance we can say

$f = kx$

$k = \frac{f}{x}$

now here we will we say that energy stored in the spring will convert into kinetic energy

$\frac{1}{2} kx^2 = \frac{1}{2}mv^2$

$\frac{f}{x} (x^2} = mv^2$

now solving above equation we will have

$v =\sqrt{ \frac{fx}{m}}$

PART 2)

now for half of the extension again we can use energy conservation

$\frac{1}{2}kx^2 - \frac{1}{2}k(x/2)^2 = \frac{1}{2} mv^2$

$\frac{3}{4}kx^2 = mv^2$

$\frac{3}{4}fx = mv^2$

now the speed is given as

$v = \sqrt{\frac{3fx}{4m}}$