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3 years ago

A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Physics

2 answers:

MaRussiya [10]3 years ago

7 0

The work required to strecth the spring is about 2.0 Joules

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<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

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

<em>where:</em>

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

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<u>Given:</u>

spring constant = k = 2500 N/m

extension = x = 4.00 cm = 4.00 × 10⁻² m

<u>Asked:</u>

work required = W = ?

<u>Solution:</u>

<em>We will calculate the </em><em>work</em><em> required to stretch the spring as follows:</em>

$W = E_p$

$W = \frac{1}{2} k x^2$

$W = \frac{1}{2} \times 2500 \times ( 4.00 \times 10^{-2} )^2$

$W = 1250 \times ( 16.00 \times 10^{-4} )$

$W = 20000 \times 10^{-4}$

$W = 2.0 \times 10^4 \times 10^{-4}$

$W = 2.0 \texttt{ J}$

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<h3>Conclusion:</h3>

The work required to strecth the spring is about 2.0 Joules

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<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840
Light Ideal Spring Speed : brainly.com/question/13050880

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

Yuri [45]3 years ago

5 0

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.

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Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

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$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

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